Construction of Similar Triangles Questions in English

(N/A) Given a triangle $ABC$,we are required to construct another triangle whose sides are $\frac{3}{4}$ of the corresponding sides of the triangle $ABC$.
Steps of Construction:
$1.$ Draw any ray $BX$ making an acute angle with $BC$ on the side opposite to the vertex $A$.
$2.$ Locate $4$ (the greater of $3$ and $4$ in $\frac{3}{4}$) points $B_1, B_2, B_3$ and $B_4$ on $BX$ so that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4$.
$3.$ Join $B_4C$ and draw a line through $B_3$ (the $3rd$ point,$3$ being smaller of $3$ and $4$ in $\frac{3}{4}$) parallel to $B_4C$ to intersect $BC$ at $C'$.
$4.$ Draw a line through $C'$ parallel to the line $CA$ to intersect $BA$ at $A'$.
Then,$\Delta A'BC'$ is the required triangle.
Let us now see how this construction gives the required triangle.
Since $B_3C' \parallel B_4C$,by Basic Proportionality Theorem,$\frac{BC'}{C'C} = \frac{BB_3}{B_3B_4} = \frac{3}{1}$.
Therefore,$\frac{BC}{BC'} = \frac{BC' + C'C}{BC'} = 1 + \frac{C'C}{BC'} = 1 + \frac{1}{3} = \frac{4}{3}$,i.e.,$\frac{BC'}{BC} = \frac{3}{4}$.
Also,$C'A' \parallel CA$. Therefore,$\Delta A'BC' \sim \Delta ABC$.
So,$\frac{A'B}{AB} = \frac{A'C'}{AC} = \frac{BC'}{BC} = \frac{3}{4}$.

(N/A) Given a triangle $ABC$,we are required to construct a triangle whose sides are $\frac{5}{3}$ of the corresponding sides of $\Delta ABC$.
Steps of Construction:
$1.$ Draw any ray $BX$ making an acute angle with $BC$ on the side opposite to the vertex $A$.
$2.$ Locate $5$ points (the greater of $5$ and $3$ in $\frac{5}{3}$) $B_{1}, B_{2}, B_{3}, B_{4}$ and $B_{5}$ on $BX$ so that $BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4} = B_{4}B_{5}$.
$3.$ Join $B_{3}$ (the $3$rd point,$3$ being smaller of $3$ and $5$ in $\frac{5}{3}$) to $C$ and draw a line through $B_{5}$ parallel to $B_{3}C$,intersecting the extended line segment $BC$ at $C^{\prime}$.
$4.$ Draw a line through $C^{\prime}$ parallel to $CA$ intersecting the extended line segment $BA$ at $A^{\prime}$.
Then $\Delta A^{\prime}BC^{\prime}$ is the required triangle.
For justification of the construction,note that $\Delta ABC \sim \Delta A^{\prime}BC^{\prime}$.
Therefore,$\frac{AB}{A^{\prime}B} = \frac{AC}{A^{\prime}C^{\prime}} = \frac{BC}{BC^{\prime}}$.
But,$\frac{BC}{BC^{\prime}} = \frac{BB_{3}}{BB_{5}} = \frac{3}{5}$.
So,$\frac{BC^{\prime}}{BC} = \frac{5}{3}$,and therefore,$\frac{A^{\prime}B}{AB} = \frac{A^{\prime}C^{\prime}}{AC} = \frac{BC^{\prime}}{BC} = \frac{5}{3}$.

(N/A) $1.$ Draw a line segment $AB = 4 \, cm$. Taking point $A$ as centre,draw an arc of $5 \, cm$ radius. Similarly,taking point $B$ as its centre,draw an arc of $6 \, cm$ radius. These arcs will intersect each other at point $C$. Now,$AC = 5 \, cm$ and $BC = 6 \, cm$,and $\triangle ABC$ is the required triangle.
$2.$ Draw a ray $AX$ making an acute angle with line $AB$ on the opposite side of vertex $C$.
$3.$ Locate $3$ points $A_1, A_2, A_3$ on line $AX$ such that $AA_1 = A_1A_2 = A_2A_3$.
$4.$ Join $BA_3$ and draw a line through $A_2$ parallel to $BA_3$ to intersect $AB$ at point $B'$.
$5.$ Draw a line through $B'$ parallel to the line $BC$ to intersect $AC$ at $C'$.
$\triangle AB'C'$ is the required triangle.
The construction can be justified by proving that $\frac{AB'}{AB} = \frac{B'C'}{BC} = \frac{AC'}{AC} = \frac{2}{3}$.
By construction,we have $B'C' \parallel BC$.
$\therefore \angle AB'C' = \angle ABC$ (Corresponding angles).
In $\triangle AB'C'$ and $\triangle ABC$,
$\angle AB'C' = \angle ABC$ (Proved above),
$\angle B'AC' = \angle BAC$ (Common angle).
$\triangle AB'C' \sim \triangle ABC$ ($AA$ similarity criterion).
$\Rightarrow \frac{AB'}{AB} = \frac{B'C'}{BC} = \frac{AC'}{AC}$ .......... $(1)$
In $\triangle AA_2B'$ and $\triangle AA_3B$,
$\angle A_2AB' = \angle A_3AB$ (Common angle),
$\angle AA_2B' = \angle AA_3B$ (Corresponding angles).
$\therefore \triangle AA_2B' \sim \triangle AA_3B$ ($AA$ similarity criterion).
$\Rightarrow \frac{AB'}{AB} = \frac{AA_2}{AA_3} = \frac{2}{3}$ ......... $(2)$
From equations $(1)$ and $(2)$,we obtain $\frac{AB'}{AB} = \frac{B'C'}{BC} = \frac{AC'}{AC} = \frac{2}{3}$.
This justifies the construction.

(N/A) $1.$ Draw a line segment $AB$ of $5\, cm$. Taking $A$ and $B$ as centers,draw arcs of $6\, cm$ and $7\, cm$ radius respectively. Let these arcs intersect each other at point $C$. $\triangle ABC$ is the required triangle having side lengths of $5\, cm$,$6\, cm$,and $7\, cm$ respectively.
$2.$ Draw a ray $AX$ making an acute angle with line $AB$ on the opposite side of vertex $C$.
$3.$ Locate $7$ points,$A_1, A_2, A_3, A_4, A_5, A_6, A_7$ (as $7$ is greater between $5$ and $7$) on line $AX$ such that $AA_1 = A_1A_2 = A_2A_3 = A_3A_4 = A_4A_5 = A_5A_6 = A_6A_7$.
$4.$ Join $BA_5$ and draw a line through $A_7$ parallel to $BA_5$ to intersect the extended line segment $AB$ at point $B'$.
$5.$ Draw a line through $B'$ parallel to $BC$ intersecting the extended line segment $AC$ at $C'$. $\triangle AB'C'$ is the required triangle.
Justification:
The construction can be justified by proving that $AB' = \frac{7}{5} AB$,$B'C' = \frac{7}{5} BC$,and $AC' = \frac{7}{5} AC$.
In $\triangle ABC$ and $\triangle AB'C'$:
$\angle ABC = \angle AB'C'$ (Corresponding angles)
$\angle BAC = \angle B'AC'$ (Common)
$\triangle ABC \sim \triangle AB'C'$ ($AA$ similarity criterion)
$\Rightarrow \frac{AB}{AB'} = \frac{BC}{B'C'} = \frac{AC}{AC'} = \frac{5}{7}$ (Equation $1$)
In $\triangle AA_5B$ and $\triangle AA_7B'$:
$\angle A_5AB = \angle A_7AB'$ (Common)
$\angle AA_5B = \angle AA_7B'$ (Corresponding angles)
$\triangle AA_5B \sim \triangle AA_7B'$ ($AA$ similarity criterion)
$\Rightarrow \frac{AB}{AB'} = \frac{AA_5}{AA_7} = \frac{5}{7}$ (Equation $2$)
On comparing equations $1$ and $2$,we obtain:
$\frac{AB}{AB'} = \frac{BC}{B'C'} = \frac{AC}{AC'} = \frac{5}{7}$
$\Rightarrow AB' = \frac{7}{5} AB, B'C' = \frac{7}{5} BC, AC' = \frac{7}{5} AC$.
This justifies the construction.

(N/A) Let us assume that $\triangle ABC$ is an isosceles triangle having $CA = CB$,base $AB = 8\, cm$,and altitude $AD = 4\, cm$ (where $D$ is the midpoint of $AB$).
$A$ $\triangle AB'C'$ whose sides are $\frac{3}{2}$ times those of $\triangle ABC$ can be drawn as follows:
$1.$ Draw a line segment $AB = 8\, cm$. Construct the perpendicular bisector of $AB$ to find the midpoint $D$. Let the perpendicular line be $OO'$.
$2.$ Taking $D$ as the centre,draw an arc of radius $4\, cm$ on the perpendicular line to intersect at point $C$. Join $AC$ and $BC$ to form the isosceles $\triangle ABC$.
$3.$ Draw a ray $AX$ making an acute angle with $AB$ on the side opposite to vertex $C$.
$4.$ Locate $3$ points $A_1, A_2,$ and $A_3$ on $AX$ such that $AA_1 = A_1A_2 = A_2A_3$.
$5.$ Join $BA_2$. Draw a line through $A_3$ parallel to $BA_2$ to intersect the extended line segment $AB$ at $B'$.
$6.$ Draw a line through $B'$ parallel to $BC$ to intersect the extended line segment $AC$ at $C'$. $\triangle AB'C'$ is the required triangle.
Justification:
Since $A_2B \parallel A_3B'$,by Basic Proportionality Theorem in $\triangle AA_3B'$,we have $\frac{AB}{AB'} = \frac{AA_2}{AA_3} = \frac{2}{3}$. Thus,$AB' = \frac{3}{2}AB$.
Since $BC \parallel B'C'$,$\triangle ABC \sim \triangle AB'C'$. Therefore,$\frac{AB}{AB'} = \frac{BC}{B'C'} = \frac{AC}{AC'} = \frac{2}{3}$,which implies $B'C' = \frac{3}{2}BC$ and $AC' = \frac{3}{2}AC$. This justifies the construction.

(A) $\angle B = 45^{\circ}, \angle A = 105^{\circ}$
The sum of all interior angles in a triangle is $180^{\circ}$.
$\angle A + \angle B + \angle C = 180^{\circ}$
$105^{\circ} + 45^{\circ} + \angle C = 180^{\circ}$
$\angle C = 180^{\circ} - 150^{\circ} = 30^{\circ}$
The required triangle can be drawn as follows:
$1.$ Draw a $\triangle ABC$ with side $BC = 7 \, cm$,$\angle B = 45^{\circ}$,$\angle C = 30^{\circ}$.
$2.$ Draw a ray $BX$ making an acute angle with $BC$ on the opposite side of vertex $A$.
$3.$ Locate $4$ points $B_1, B_2, B_3, B_4$ on $BX$ such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4$.
$4.$ Join $B_3C$. Draw a line through $B_4$ parallel to $B_3C$ intersecting the extended $BC$ at $C'$.
$5.$ Through $C'$,draw a line parallel to $AC$ intersecting the extended line segment $BA$ at $A'$. $\triangle A'BC'$ is the required triangle.
Justification:
In $\triangle ABC$ and $\triangle A'BC'$,
$\angle ABC = \angle A'BC'$ (Common)
$\angle ACB = \angle A'C'B$ (Corresponding angles)
$\triangle ABC \sim \triangle A'BC'$ ($AA$ similarity criterion)
$\Rightarrow \frac{AB}{A'B} = \frac{BC}{BC'} = \frac{AC}{A'C'} \dots(1)$
In $\triangle BB_3C$ and $\triangle BB_4C'$,
$\angle B_3BC = \angle B_4BC'$ (Common)
$\angle BB_3C = \angle BB_4C'$ (Corresponding angles)
$\triangle BB_3C \sim \triangle BB_4C'$ ($AA$ similarity criterion)
$\Rightarrow \frac{BC}{BC'} = \frac{BB_3}{BB_4} = \frac{3}{4} \dots(2)$
From $(1)$ and $(2)$,we get $\frac{AB}{A'B} = \frac{BC}{BC'} = \frac{AC}{A'C'} = \frac{3}{4}$.
Thus,$A'B = \frac{4}{3} AB$,$BC' = \frac{4}{3} BC$,and $A'C' = \frac{4}{3} AC$. This justifies the construction.

(N/A) It is given that sides other than hypotenuse are of lengths $4 \,cm$ and $3 \,cm$. Clearly,these will be perpendicular to each other.
The required triangle can be drawn as follows:
$1.$ Draw a line segment $AB = 4 \,cm$. Draw a ray $AX$ making $90^{\circ}$ with it at $A$.
$2.$ Draw an arc of $3 \,cm$ radius while taking $A$ as its centre to intersect the ray at $C$. Join $BC$. $\triangle ABC$ is the required triangle.
$3.$ Draw a ray $AY$ making an acute angle with $AB$,opposite to vertex $C$.
$4.$ Locate $5$ points $A_1, A_2, A_3, A_4, A_5$ on $AY$ such that $AA_1 = A_1A_2 = A_2A_3 = A_3A_4 = A_4A_5$.
$5.$ Join $A_3B$. Draw a line through $A_5$ parallel to $A_3B$ intersecting the extended line segment $AB$ at $B'$.
$6.$ Through $B'$,draw a line parallel to $BC$ intersecting the extended line segment $AC$ at $C'$. $\triangle AB'C'$ is the required triangle.
Justification:
The construction can be justified by proving that $AB' = \frac{5}{3} AB, B'C' = \frac{5}{3} BC, AC' = \frac{5}{3} AC$.
In $\triangle ABC$ and $\triangle AB'C'$,
$\angle ABC = \angle AB'C'$ (Corresponding angles)
$\angle BAC = \angle B'AC'$ (Common)
$\therefore \triangle ABC \sim \triangle AB'C'$ ($AA$ similarity criterion)
$\Rightarrow \frac{AB}{AB'} = \frac{BC}{B'C'} = \frac{AC}{AC'}$ .......$(1)$
In $\triangle AA_3B$ and $\triangle AA_5B'$,
$\angle A_3AB = \angle A_5AB'$ (Common)
$\angle AA_3B = \angle AA_5B'$ (Corresponding angles)
$\therefore \triangle AA_3B \sim \triangle AA_5B'$ ($AA$ similarity criterion)
$\Rightarrow \frac{AB}{AB'} = \frac{AA_3}{AA_5} = \frac{3}{5}$ .........$(2)$
On comparing equations $(1)$ and $(2)$,we obtain
$\frac{AB}{AB'} = \frac{BC}{B'C'} = \frac{AC}{AC'} = \frac{3}{5}$
$AB' = \frac{5}{3} AB, B'C' = \frac{5}{3} BC, AC' = \frac{5}{3} AC$
This justifies the construction.

(N/A) $\triangle A'BC'$ whose sides are $\frac{3}{4}$ of the corresponding sides of $\triangle ABC$ can be drawn as follows:
$1.$ Draw a $\triangle ABC$ with side $BC = 6 \, cm$,$AB = 5 \, cm$ and $\angle ABC = 60^{\circ}$.
$2.$ Draw a ray $BX$ making an acute angle with $BC$ on the opposite side of vertex $A$.
$3.$ Locate $4$ points (as $4$ is the greater number in $\frac{3}{4}$),$B_1, B_2, B_3, B_4$,on line segment $BX$ such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4$.
$4.$ Join $B_4C$ and draw a line through $B_3$ parallel to $B_4C$ intersecting $BC$ at $C'$.
$5.$ Draw a line through $C'$ parallel to $AC$ intersecting $AB$ at $A'$. $\triangle A'BC'$ is the required triangle.
Justification:
The construction can be justified by proving $A'B = \frac{3}{4} AB$,$BC' = \frac{3}{4} BC$,$A'C' = \frac{3}{4} AC$.
In $\triangle A'BC'$ and $\triangle ABC$:
$\angle A'C'B = \angle ACB$ (Corresponding angles)
$\angle A'BC' = \angle ABC$ (Common)
$\therefore \triangle A'BC' \sim \triangle ABC$ ($AA$ similarity criterion)
$\Rightarrow \frac{A'B}{AB} = \frac{BC'}{BC} = \frac{A'C'}{AC}$ ........$(1)$
In $\triangle BB_3C'$ and $\triangle BB_4C$:
$\angle B_3BC' = \angle B_4BC$ (Common)
$\angle BB_3C' = \angle BB_4C$ (Corresponding angles)
$\therefore \triangle BB_3C' \sim \triangle BB_4C$ ($AA$ similarity criterion)
$\Rightarrow \frac{BC'}{BC} = \frac{BB_3}{BB_4} = \frac{3}{4}$ ........$(2)$
From equations $(1)$ and $(2)$,we obtain:
$\frac{A'B}{AB} = \frac{BC'}{BC} = \frac{A'C'}{AC} = \frac{3}{4}$
$A'B = \frac{3}{4} AB, BC' = \frac{3}{4} BC, A'C' = \frac{3}{4} AC$.
This justifies the construction.

(B) To construct a triangle similar to a given $\triangle ABC$ with a scale factor of $\frac{3}{7}$,we follow these steps:
$1$. Draw a ray $BX$ making an acute angle with $BC$.
$2$. Locate $7$ points $(B_{1}, B_{2}, B_{3}, B_{4}, B_{5}, B_{6}, B_{7})$ on $BX$ at equal distances because the denominator of the fraction $\frac{3}{7}$ is $7$.
$3$. The next step is to join the last point,$B_{7}$,to $C$ to establish the base line for the construction.

(C) To construct a triangle similar to a given triangle with a scale factor of $\frac{m}{n}$,where $m$ and $n$ are positive integers,the number of points to be marked on the ray $BX$ is equal to the greater of $m$ and $n$.
In this problem,the scale factor is $\frac{8}{5}$.
Comparing the numerator $m = 8$ and the denominator $n = 5$,the greater value is $8$.
Therefore,the minimum number of points to be located at equal distances on ray $BX$ is $8$.

(B) False.
To construct a triangle similar to $\triangle ABC$ with a scale factor of $\frac{7}{3}$,we must divide the ray $BX$ into $7$ equal parts (since $7 > 3$).
Steps of construction:
$1.$ Draw a ray $BX$ making an acute angle with $BC$.
$2.$ Locate $7$ points $B_{1}, B_{2}, \dots, B_{7}$ on $BX$ such that $BB_{1} = B_{1}B_{2} = \dots = B_{6}B_{7}$.
$3.$ Join $B_{3}$ to $C$ (since the denominator is $3$).
$4.$ Draw a line through $B_{7}$ parallel to $B_{3}C$ intersecting the extended line segment $BC$ at $C'$.
$5.$ Draw a line through $C'$ parallel to $AC$ intersecting the extended line segment $BA$ at $A'$.
The given statement claims that $B_{6}C'$ is drawn parallel to $B_{3}C$,which is incorrect because the scale factor is $\frac{7}{3}$,requiring the parallel line to be drawn from $B_{7}$,not $B_{6}$.

(A) $1$. First,draw an equilateral triangle $ABC$ where $AB = BC = CA = 4\, cm$.
$2$. Draw a ray $AX$ making an acute angle with $AB$ on the side opposite to vertex $C$.
$3$. Locate $5$ points $A_1, A_2, A_3, A_4, A_5$ on $AX$ such that $AA_1 = A_1A_2 = A_2A_3 = A_3A_4 = A_4A_5$.
$4$. Join $A_5$ to $B$. Draw a line through $A_3$ parallel to $A_5B$ intersecting $AB$ at $B'$.
$5$. Draw a line through $B'$ parallel to $BC$ intersecting $AC$ at $C'$.
$6$. The triangle $AB'C'$ is the required triangle similar to $\triangle ABC$ with a scale factor of $\frac{3}{5}$.
$7$. Since the ratio of corresponding sides is equal,the angles remain the same ($60^{\circ}$ each). Therefore,the new triangle is also equilateral.

(A) Steps of construction:
$1.$ Draw a line segment $BC = 12 \, cm$.
$2.$ From point $B$,draw a line $AB = 5 \, cm$ perpendicular to $BC$ such that $\angle B = 90^{\circ}$.
$3.$ Join $AC$. Thus,$\triangle ABC$ is the given right triangle.
$4.$ From $B$,draw an acute angle $\angle CBY$ downwards.
$5.$ On ray $BY$,mark three points $B_1, B_2$ and $B_3$ such that $BB_1 = B_1B_2 = B_2B_3$.
$6.$ Join $B_3C$.
$7.$ From point $B_2$,draw $B_2N \parallel B_3C$ to intersect $BC$ at $N$.
$8.$ From point $N$,draw $NM \parallel CA$ to intersect $BA$ at $M$. Thus,$\triangle MBN$ is the required triangle.
$9.$ Since $NM \parallel CA$ and $BC$ is a transversal,the corresponding angles are equal. Therefore,$\angle MNB = \angle ACB$. Since $\angle B = 90^{\circ}$ in both triangles,$\triangle MBN$ is also a right-angled triangle at $B$.

(A) $1.$ Draw a line segment $BC = 6 \, cm$.
$2.$ Taking $B$ and $C$ as centers,draw two arcs of radii $4 \, cm$ and $5 \, cm$ respectively,intersecting each other at $A$.
$3.$ Join $BA$ and $CA$. $\triangle ABC$ is the required triangle.
$4.$ From $B$,draw any ray $BX$ downwards making an acute angle with $BC$.
$5.$ Mark five points $B_1, B_2, B_3, B_4,$ and $B_5$ on $BX$ such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4 = B_4B_5$.
$6.$ Join $B_3C$. From $B_5$,draw $B_5M \parallel B_3C$ intersecting the extended line segment $BC$ at $M$.
$7.$ From point $M$,draw $MN \parallel CA$ intersecting the extended line segment $BA$ at $N$. Then,$\triangle NBM$ is the required triangle whose sides are $\frac{5}{3}$ times the corresponding sides of $\triangle ABC$.

(N/A) First,draw the rhombus $ABCD$ in which $AB = 4 \, cm$ and $\angle ABC = 60^{\circ}$ as shown in the figure,and join $AC$.
Construct the triangle $AB'C'$ similar to $\triangle ABC$ with a scale factor of $\frac{2}{3}$ as instructed in the Mathematics Textbook for Class $X$.
Finally,draw the line segment $C'D'$ parallel to $CD$ such that $D'$ lies on $AD$.
Since $\triangle AB'C' \sim \triangle ABC$,we have $\frac{AB'}{AB} = \frac{AC'}{AC} = \frac{B'C'}{BC} = \frac{2}{3}$.
Since $C'D' \parallel CD$,$\triangle AD'C' \sim \triangle ADC$. Thus,$\frac{AD'}{AD} = \frac{AC'}{AC} = \frac{C'D'}{CD} = \frac{2}{3}$.
Since $ABCD$ is a rhombus,$AB = BC = CD = AD$.
Therefore,$AB' = B'C' = C'D' = AD' = \frac{2}{3} AB$.
Since all four sides of the quadrilateral $AB'C'D'$ are equal,$AB'C'D'$ is a rhombus.

(A) $1.$ Draw a line segment $AB = 3 \, cm$.
$2.$ Draw a ray $BY$ making an angle $\angle ABY = 60^{\circ}$.
$3.$ With $B$ as the center and a radius of $5 \, cm$,draw an arc to cut the ray $BY$ at point $C$.
$4.$ Draw a ray $AZ$ parallel to $BY$ such that $\angle BAZ = 120^{\circ}$ (since $ABCD$ is a parallelogram,$\angle ABC + \angle BCD = 180^{\circ}$).
$5.$ With $A$ as the center and a radius of $5 \, cm$,draw an arc to cut the ray $AZ$ at point $D$.
$6.$ Join $CD$ to complete the parallelogram $ABCD$.
$7.$ Join $BD$,which is a diagonal of the parallelogram $ABCD$.
$8.$ From $B$,draw any ray $BX$ downwards making an acute angle $\angle CBX$.
$9.$ Locate $4$ points $B_1, B_2, B_3, B_4$ on $BX$ such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4$.
$10.$ Join $B_3C$ and draw a line through $B_4$ parallel to $B_3C$ intersecting the extended line segment $BC$ at $C'$.
$11.$ From point $C'$,draw $C'D' \parallel CD$ intersecting the extended line segment $BD$ at $D'$.
$12.$ Draw a line segment $D'A'$ parallel to $DA$,where $A'$ lies on the extended side $BA$.
$13.$ Yes,$A'BCD'$ is a parallelogram because by construction,$A'D' \parallel BC$ and $A'B \parallel D'C'$.

(N/A) Let $\triangle PQR$ and $\triangle ABC$ be similar triangles. The scale factor between the corresponding sides is $\frac{PQ}{AB} = \frac{8}{6} = \frac{4}{3}$.
Steps of construction:
$1.$ Draw a line segment $BC = 5 \, cm$.
$2.$ Taking $B$ and $C$ as centers,draw two arcs of radius $6 \, cm$ intersecting each other at $A$.
$3.$ Join $BA$ and $CA$. Thus,$\triangle ABC$ is the required isosceles triangle.
$4.$ From $B$,draw any ray $BX$ making an acute $\angle CBX$.
$5.$ Locate four points $B_1, B_2, B_3,$ and $B_4$ on $BX$ such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4$.
$6.$ Join $B_3C$ and from $B_4$,draw a line $B_4R \parallel B_3C$ intersecting the extended line segment $BC$ at $R$.
$7.$ From point $R$,draw $RP \parallel CA$ meeting the extended line segment $BA$ at $P$.
Then,$\triangle PBR$ is the required triangle.
Justification:
Since $B_4R \parallel B_3C$ (by construction),
$\therefore \frac{BC}{CR} = \frac{3}{1}$.
Now,$\frac{BR}{BC} = \frac{BC + CR}{BC} = 1 + \frac{CR}{BC} = 1 + \frac{1}{3} = \frac{4}{3}$.
Also,$RP \parallel CA$,therefore $\triangle ABC \sim \triangle PBR$.
And $\frac{PB}{AB} = \frac{RP}{CA} = \frac{BR}{BC} = \frac{4}{3}$.
Hence,the new triangle is similar to the given triangle with sides $\frac{4}{3}$ times the corresponding sides of the isosceles $\triangle ABC$.

(N/A) Steps of construction:
$1.$ Draw a line segment $AB = 5 \, cm$.
$2.$ At point $B$,construct $\angle ABY = 60^{\circ}$ and cut off $BC = 6 \, cm$ on $BY$.
$3.$ Join $AC$. Thus,$\triangle ABC$ is the required triangle.
$4.$ Draw any ray $AX$ making an acute angle with $AB$ on the side opposite to vertex $C$.
$5.$ Locate $7$ points $B_1, B_2, B_3, B_4, B_5, B_6, B_7$ on $AX$ such that $AB_1 = B_1B_2 = B_2B_3 = B_3B_4 = B_4B_5 = B_5B_6 = B_6B_7$.
$6.$ Join $B_7B$. Through $B_5$,draw a line parallel to $B_7B$ intersecting $AB$ at $M$.
$7.$ Through $M$,draw a line parallel to $BC$ intersecting $AC$ at $N$.
Then,$\triangle AMN$ is the required triangle whose sides are $\frac{5}{7}$ of the corresponding sides of $\triangle ABC$.
Justification:
Since $MN \parallel BC$,by Basic Proportionality Theorem,$\triangle AMN \sim \triangle ABC$.
By construction,$B_5M \parallel B_7B$. In $\triangle ABB_7$,by Thales theorem:
$\frac{AM}{AB} = \frac{AB_5}{AB_7} = \frac{5}{7}$.
Since $\triangle AMN \sim \triangle ABC$,the ratio of their corresponding sides is $\frac{AM}{AB} = \frac{AN}{AC} = \frac{MN}{BC} = \frac{5}{7}$.

(N/A) Steps of construction:
$1.$ Draw a line segment $BC = 6 \, cm$.
$2.$ Taking $B$ and $C$ as centers,draw two arcs of radii $4 \, cm$ and $9 \, cm$ respectively,intersecting each other at $A$.
$3.$ Join $BA$ and $CA$. $\triangle ABC$ is the required triangle.
$4.$ From $B$,draw any ray $BX$ downwards making an acute angle.
$5.$ Mark three points $B_1, B_2, B_3$ on $BX$ such that $BB_1 = B_1B_2 = B_2B_3$.
$6.$ Join $B_2C$ and from $B_3$,draw $B_3M \parallel B_2C$ intersecting the extended line segment $BC$ at $M$.
$7.$ From point $M$,draw $MN \parallel CA$ intersecting the extended line segment $BA$ at $N$.
Then,$\triangle NBM$ is the required triangle whose sides are $\frac{3}{2}$ times the corresponding sides of $\triangle ABC$.
The two triangles are not congruent because congruent triangles must have the same shape and size. Here,the triangles are similar (same shape),but their sizes are different.
Justification:
Since $B_3M \parallel B_2C$,by Basic Proportionality Theorem:
$\frac{BC}{CM} = \frac{BB_2}{B_2B_3} = \frac{2}{1}$.
Now,$\frac{BM}{BC} = \frac{BC + CM}{BC} = 1 + \frac{CM}{BC} = 1 + \frac{1}{2} = \frac{3}{2}$.
Also,$MN \parallel CA$,therefore $\triangle ABC \sim \triangle NBM$.
Thus,$\frac{NB}{AB} = \frac{NM}{AC} = \frac{BM}{BC} = \frac{3}{2}$.

(N/A) Data: Construct $\Delta ABC$ in which $AB = 4 \, cm$,$BC = 7 \, cm$,and $AC = 5 \, cm$.
To construct: Construct $\Delta BPQ$ similar to $\Delta ABC$ such that the ratio of their corresponding sides is $2:3$.
Steps of construction:
$(1)$ Construct $\Delta ABC$ in which $AB = 4 \, cm$,$BC = 7 \, cm$,and $AC = 5 \, cm$.
$(2)$ In the half-plane of $\overleftrightarrow{BC}$ that does not contain $A$,draw a ray $\overrightarrow{BX}$ such that $\angle CBX$ is an acute angle.
$(3)$ With an appropriate radius and center $B$,draw an arc to intersect $\overrightarrow{BX}$ at $B_1$. Similarly,with center $B_1$ and the same radius,draw an arc to intersect $\overrightarrow{BX}$ at $B_2$ such that $B-B_1-B_2$. Again,with the same radius and center $B_2$,draw an arc to intersect $\overrightarrow{BX}$ at $B_3$ such that $B_1-B_2-B_3$.
$(4)$ Draw the line segment $\overline{B_3C}$.
$(5)$ Through $B_2$,draw a line parallel to $\overline{B_3C}$ to intersect $\overline{BC}$ at $P$.
$(6)$ Through $P$,draw a line parallel to $\overline{CA}$ to intersect $\overline{AB}$ at $Q$.
Thus,$\Delta BPQ$ is the required triangle.
Justification: $\overleftrightarrow{BX}$ and $\overleftrightarrow{BC}$ are transversals to $\overleftrightarrow{B_3C} \parallel \overleftrightarrow{B_2P}$.
Therefore,$BP:BC = BB_2:BB_3 = 2:3$.
Similarly,$\overleftrightarrow{BC}$ and $\overleftrightarrow{BA}$ are transversals to $\overleftrightarrow{CA} \parallel \overleftrightarrow{PQ}$.
Therefore,$BQ:BA = BP:BC = 2:3$.

(N/A) Data: Construct $\Delta PQR$ with $m \angle P = 60^{\circ}, m \angle Q = 45^{\circ}$ and $PQ = 6 \text{ cm}$.
To construct: Construct $\Delta PBC$ similar to $\Delta PQR$ such that the ratio of their corresponding sides is $5:3$.
Steps of construction:
$(1)$ Construct $\Delta PQR$ with $m \angle P=60^{\circ}, m \angle Q=45^{\circ}$ and $PQ = 6 \text{ cm}$.
$(2)$ In the half-plane of $\overleftrightarrow{PQ}$ that does not contain $R$,draw a ray $\overrightarrow{PZ}$ such that $\angle QPZ$ is an acute angle.
$(3)$ With some appropriate radius and center $P$,draw an arc to intersect $\overrightarrow{PZ}$ at $P_1$. With the same radius and center $P_1$,draw an arc to intersect $\overrightarrow{PZ}$ at $P_2$ such that $P-P_1-P_2$. Similarly,with the same radius and center $P_k$,draw an arc to intersect $\overrightarrow{PZ}$ at $P_{k+1}$ such that $P_{k-1}-P_k-P_{k+1}$,where $k=2, 3, 4$.
$(4)$ Draw $\overline{P_3Q}$.
$(5)$ Through $P_5$,draw a ray parallel to $\overline{P_3Q}$ to intersect the extended ray $\overrightarrow{PQ}$ at $B$.
$(6)$ Through $B$,draw a ray parallel to $\overline{QR}$ to intersect the extended ray $\overrightarrow{PR}$ at $C$.
Thus,$\Delta PBC$ is the required triangle.
Justification: In $\Delta PP_5B$,$P-P_3-P_5$,$P-Q-B$ and $\overline{P_3Q} \parallel \overline{P_5B}$.
$\therefore \frac{PB}{PQ} = \frac{PP_5}{PP_3} = \frac{5}{3}$.
Similarly,in $\Delta PBC$,$P-Q-B$,$P-R-C$ and $\overline{QR} \parallel \overline{BC}$.
$\therefore \frac{PC}{PR} = \frac{PB}{PQ} = \frac{5}{3}$.

(N/A) Steps of construction:
$1$. Draw a line segment $BC = 6 \, cm$.
$2$. With $B$ as the center and radius $3 \, cm$,draw an arc. With $C$ as the center and radius $5 \, cm$,draw another arc intersecting the first arc at $A$.
$3$. Join $AB$ and $AC$ to obtain $\Delta ABC$.
$4$. Draw a ray $BX$ making an acute angle with $BC$ on the side opposite to vertex $A$.
$5$. Locate $4$ points $B_1, B_2, B_3, B_4$ on $BX$ such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4$.
$6$. Join $B_4C$.
$7$. Draw a line through $B_3$ parallel to $B_4C$ intersecting $BC$ at $Q$.
$8$. Draw a line through $Q$ parallel to $AC$ intersecting $AB$ at $P$.
$9$. $\Delta BPQ$ is the required triangle.

(N/A) Steps of construction:
$1$. Draw a line segment $BC = 7 \text{ cm}$.
$2$. With $B$ as the center and radius $6 \text{ cm}$,draw an arc. With $C$ as the center and radius $5 \text{ cm}$,draw another arc intersecting the first arc at $A$.
$3$. Join $AB$ and $AC$ to form $\Delta ABC$.
$4$. Draw a ray $BX$ making an acute angle with $BC$ on the side opposite to vertex $A$.
$5$. Locate $5$ points $B_1, B_2, B_3, B_4, B_5$ on $BX$ such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4 = B_4B_5$.
$6$. Join $B_5C$.
$7$. From $B_4$,draw $B_4Y \parallel B_5C$ meeting $BC$ at $Y$.
$8$. From $Y$,draw $YX \parallel CA$ meeting $AB$ at $X$. Then $\Delta BXY$ is the required triangle.

(N/A) Steps of construction:
$1$. Draw a line segment $QR = 7 \, cm$.
$2$. At point $Q$,construct an angle of $45^{\circ}$ using a protractor or compass.
$3$. At point $R$,construct an angle of $60^{\circ}$ using a compass.
$4$. Let the intersection of these two rays be point $P$. Thus,$\Delta PQR$ is formed.
$5$. Draw an acute angle $\angle RQX$ below the base $QR$.
$6$. Locate $4$ points $Q_1, Q_2, Q_3, Q_4$ on ray $QX$ such that $QQ_1 = Q_1Q_2 = Q_2Q_3 = Q_3Q_4$.
$7$. Join $Q_3$ to $R$. Draw a line through $Q_4$ parallel to $Q_3R$ intersecting the extended line segment $QR$ at $R'$.
$8$. Draw a line through $R'$ parallel to $RP$ intersecting the extended line segment $QP$ at $P'$.
$9$. $\Delta P'QR'$ is the required triangle similar to $\Delta PQR$ with a scale factor of $\frac{4}{3}$.

(N/A) Steps of construction:
$1$. Draw an equilateral triangle $\Delta XYZ$ with each side equal to $6\, cm$.
$2$. Draw a ray $XS$ making an acute angle with $XY$ at vertex $X$ on the side opposite to $Z$.
$3$. Locate $5$ points $X_1, X_2, X_3, X_4, X_5$ on ray $XS$ such that $XX_1 = X_1X_2 = X_2X_3 = X_3X_4 = X_4X_5$.
$4$. Join $X_4$ to $Y$. Draw a line through $X_5$ parallel to $X_4Y$ intersecting the extended line segment $XY$ at point $M$.
$5$. Draw a line through $M$ parallel to $YZ$ intersecting the extended line segment $XZ$ at point $N$.
$6$. Thus,$\Delta XMN$ is the required triangle similar to $\Delta XYZ$ with scale factor $\frac{5}{4}$.

(N/A) Steps of construction:
$1$. Draw a line segment $BC = 6 \text{ cm}$.
$2$. At point $B$,construct an angle of $90^{\circ}$ using a protractor or compass.
$3$. Cut an arc of $5 \text{ cm}$ on the ray from $B$ to mark point $A$. Join $AC$ to complete $\Delta ABC$.
$4$. Draw an acute angle $\angle CBX$ below $BC$.
$5$. Mark $3$ points $B_1, B_2, B_3$ on $BX$ such that $BB_1 = B_1B_2 = B_2B_3$.
$6$. Join $B_2$ to $C$. Draw a line through $B_3$ parallel to $B_2C$ intersecting the extended line $BC$ at $Q$.
$7$. Draw a line through $Q$ parallel to $AC$ intersecting the extended line $BA$ at $P$.
$8$. $\Delta BPQ$ is the required triangle.

(N/A) Steps of construction:
$1$. Draw a line segment $BC = 8 \, cm$.
$2$. At point $B$,construct an angle of $60^\circ$ using a compass and ruler.
$3$. From point $B$,cut an arc of $6 \, cm$ on the ray making $60^\circ$ to mark point $A$. Join $AC$ to complete $\Delta ABC$.
$4$. Draw an acute angle $\angle CBZ$ below the line segment $BC$.
$5$. Mark $5$ points $B_1, B_2, B_3, B_4, B_5$ on ray $BZ$ such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4 = B_4B_5$.
$6$. Join $B_5$ to $C$.
$7$. Draw a line from $B_4$ parallel to $B_5C$ intersecting $BC$ at $Y$.
$8$. Draw a line from $Y$ parallel to $AC$ intersecting $AB$ at $X$.
$9$. $\Delta BXY$ is the required triangle similar to $\Delta ABC$ with a scale factor of $\frac{4}{5}$.

(N/A) Steps of construction:
$1$. Draw a line segment $YZ = 6\, cm$.
$2$. With $Y$ as the center and radius $4\, cm$,draw an arc.
$3$. With $Z$ as the center and radius $7\, cm$,draw another arc intersecting the previous arc at point $X$.
$4$. Join $XY$ and $XZ$ to form $\Delta XYZ$.
$5$. Draw an acute angle $\angle ZYA$ at vertex $Y$ below the line $YZ$.
$6$. Mark $4$ points $Y_1, Y_2, Y_3, Y_4$ on $YA$ such that $YY_1 = Y_1Y_2 = Y_2Y_3 = Y_3Y_4$.
$7$. Join $Y_3$ to $Z$.
$8$. Draw a line through $Y_4$ parallel to $Y_3Z$ intersecting the extended line $YZ$ at $N$.
$9$. Draw a line through $N$ parallel to $XZ$ intersecting the extended line $YX$ at $M$.
$10$. $\Delta YMN$ is the required triangle.