Some Properties of a Triangle Questions in English

(D) The Sriyantra is a complex sacred geometry pattern used in the Atharva Veda. It is composed of $9$ interwoven isosceles triangles. These triangles are arranged in such a way that they form a total of $43$ smaller triangles,but the primary interwoven structure consists of $9$ triangles.

(C) In $\Delta TQR$,since $QT \perp PR$,$\angle QTR = 90^o$.
By the angle sum property of a triangle,the sum of angles in $\Delta TQR$ is $180^o$:
$\angle QTR + \angle TQR + \angle TRQ = 180^o$
$90^o + 40^o + x = 180^o$
$130^o + x = 180^o$
$x = 180^o - 130^o = 50^o$
Now,consider $\Delta P S R$. The exterior angle $y$ of $\Delta P S R$ is equal to the sum of the two interior opposite angles:
$y = \angle SPR + \angle PRS$
$y = 30^o + x$
$y = 30^o + 50^o = 80^o$
Thus,$x = 50^o$ and $y = 80^o$.

(B) In $\Delta XYZ$,the sum of angles is $180^o$.
$\angle XYZ + \angle YZX + \angle ZXY = 180^o$
Given $\angle XYZ = 54^o$ and $\angle ZXY = 62^o$.
$54^o + \angle YZX + 62^o = 180^o$
$\angle YZX = 180^o - 116^o = 64^o$.
Since $YO$ and $ZO$ are bisectors of $\angle XYZ$ and $\angle XZY$ respectively:
$\angle OZY = \frac{1}{2} \angle XZY = \frac{1}{2} (64^o) = 32^o$.
$\angle OYZ = \frac{1}{2} \angle XYZ = \frac{1}{2} (54^o) = 27^o$.
In $\Delta OYZ$,the sum of angles is $180^o$:
$\angle YOZ + \angle OYZ + \angle OZY = 180^o$
$\angle YOZ + 27^o + 32^o = 180^o$
$\angle YOZ = 180^o - 59^o = 121^o$.
Thus,$\angle OZY = 32^o$ and $\angle YOZ = 121^o$.

(C) Let the angles of $\triangle ABC$ be $\angle A, \angle B,$ and $\angle C$.
Given that $\angle A = \angle B + \angle C$ ...... $(1)$
In any $\triangle ABC$,the sum of all interior angles is $180^{\circ}$,so $\angle A + \angle B + \angle C = 180^{\circ}$ ...... $(2)$
[Angle sum property of a triangle]
Substituting equation $(1)$ into equation $(2)$,we get:
$\angle A + (\angle B + \angle C) = 180^{\circ}$
$\angle A + \angle A = 180^{\circ}$
$2\angle A = 180^{\circ}$
$\angle A = 90^{\circ}$
Since one angle is $90^{\circ}$,the triangle is a right triangle.

(D) The exterior angle of a triangle is equal to the sum of its two interior opposite angles.
Let each of the two equal interior opposite angles be $x$.
Given that the exterior angle is $105^{\circ}$.
Therefore,$x + x = 105^{\circ}$.
$2x = 105^{\circ}$.
$x = \frac{105^{\circ}}{2} = 52 \frac{1}{2}^{\circ}$.
Thus,each of the equal angles is $52 \frac{1}{2}^{\circ}$.

(A) Let the angles of the triangle be $5x$,$3x$,and $7x$.
As the sum of the angles of a triangle is $180^{\circ}$,we have:
$5x + 3x + 7x = 180^{\circ}$
$15x = 180^{\circ}$
$x = 180^{\circ} / 15 = 12^{\circ}$
Therefore,the angles of the triangle are:
$5 \times 12^{\circ} = 60^{\circ}$
$3 \times 12^{\circ} = 36^{\circ}$
$7 \times 12^{\circ} = 84^{\circ}$
Since each angle of the triangle is less than $90^{\circ}$,all the angles are acute.
Therefore,the triangle is an acute angled triangle.

(B) Let the triangle be $\triangle ABC$ with $\angle A = 130^{\circ}.$
In any triangle,the sum of all interior angles is $180^{\circ},$ so $\angle B + \angle C = 180^{\circ} - 130^{\circ} = 50^{\circ}.$
Let $OB$ and $OC$ be the bisectors of $\angle B$ and $\angle C$ respectively.
Then $\angle OBC = \frac{1}{2} \angle B$ and $\angle OCB = \frac{1}{2} \angle C.$
In $\triangle OBC,$ the sum of angles is $180^{\circ},$ so $\angle BOC = 180^{\circ} - (\angle OBC + \angle OCB).$
Substituting the values,$\angle BOC = 180^{\circ} - \frac{1}{2}(\angle B + \angle C).$
$\angle BOC = 180^{\circ} - \frac{1}{2}(50^{\circ}) = 180^{\circ} - 25^{\circ} = 155^{\circ}.$

(A) Given that: The ratio of the angles of a triangle is $2: 4: 3$.
Let the angles of the triangle be $\angle A, \angle B$,and $\angle C$.
Therefore,$\angle A = 2x, \angle B = 4x$,and $\angle C = 3x$.
In $\triangle ABC$,the sum of the angles is $\angle A + \angle B + \angle C = 180^{\circ}$.
[Since the sum of the interior angles of a triangle is $180^{\circ}$].
Substituting the values: $2x + 4x + 3x = 180^{\circ} \Rightarrow 9x = 180^{\circ} \Rightarrow x = 180^{\circ} / 9 = 20^{\circ}$.
Calculating the angles:
$\angle A = 2x = 2 \times 20^{\circ} = 40^{\circ}$.
$\angle B = 4x = 4 \times 20^{\circ} = 80^{\circ}$.
$\angle C = 3x = 3 \times 20^{\circ} = 60^{\circ}$.
Comparing the angles $40^{\circ}, 80^{\circ}$,and $60^{\circ}$,the smallest angle is $40^{\circ}$.

(B) No,a triangle cannot have all angles less than $60^{\circ}$.
If all three angles of a triangle are less than $60^{\circ}$,then their sum will be less than $60^{\circ} + 60^{\circ} + 60^{\circ} = 180^{\circ}$.
However,the sum of all interior angles of a triangle is always equal to $180^{\circ}$ (Angle Sum Property of a triangle).
Therefore,it is impossible for all angles to be less than $60^{\circ}$.

(NO) An angle whose measure is more than $90^{\circ}$ but less than $180^{\circ}$ is called an obtuse angle.
$A$ triangle cannot have two obtuse angles because the sum of all the interior angles of a triangle is always exactly $180^{\circ}$.
If a triangle had two obtuse angles,the sum of these two angles alone would be greater than $90^{\circ} + 90^{\circ} = 180^{\circ}$.
Since this sum exceeds the total angle sum property of a triangle $(180^{\circ})$,it is impossible for a triangle to have two obtuse angles.

(A) The sum of the interior angles of any triangle must be exactly $180^{\circ}$.
In the given case,the sum of the angles is $45^{\circ} + 64^{\circ} + 72^{\circ} = 181^{\circ}$.
Since the sum of the angles is not equal to $180^{\circ}$,it is impossible to construct such a triangle.
Therefore,$0$ triangles can be drawn.

Given: Ratio of angles is $2: 3: 4$.
To find: Angles of the triangle.
Let the angles of the triangle be $\angle A, \angle B,$ and $\angle C$.
Therefore,$\angle A = 2x$,$\angle B = 3x$,and $\angle C = 4x$.
In $\triangle ABC$,the sum of angles is $\angle A + \angle B + \angle C = 180^{\circ}$ (Angle Sum Property of a triangle).
Substituting the values: $2x + 3x + 4x = 180^{\circ}$.
$9x = 180^{\circ} \Rightarrow x = 180^{\circ} / 9 = 20^{\circ}$.
Therefore,$\angle A = 2 \times 20^{\circ} = 40^{\circ}$,$\angle B = 3 \times 20^{\circ} = 60^{\circ}$,and $\angle C = 4 \times 20^{\circ} = 80^{\circ}$.
Hence,the angles of the triangle are $40^{\circ}, 60^{\circ},$ and $80^{\circ}$.

(N/A) Let us consider a triangle $PQR$ where $\angle 1, \angle 2$,and $\angle 3$ are the interior angles of $\Delta PQR$.
We need to prove that $\angle 1 + \angle 2 + \angle 3 = 180^{\circ}$.
Draw a line $XPY$ parallel to $QR$ passing through the vertex $P$.
Since $XPY$ is a straight line,the sum of angles on it at point $P$ is $180^{\circ}$,so $\angle 4 + \angle 1 + \angle 5 = 180^{\circ}$.
Since $XPY \parallel QR$ and $PQ, PR$ are transversals,the alternate interior angles are equal:
$\angle 4 = \angle 2$ and $\angle 5 = \angle 3$.
Substituting these values into the equation $\angle 4 + \angle 1 + \angle 5 = 180^{\circ}$,we get:
$\angle 2 + \angle 1 + \angle 3 = 180^{\circ}$,which means $\angle 1 + \angle 2 + \angle 3 = 180^{\circ}$.
Thus,the sum of the three angles of a triangle is $180^{\circ}$.

(N/A) In $\Delta ABC$,we have
$AB = AC$ (Given,since $\Delta ABC$ is an isosceles triangle)
Since angles opposite to equal sides are equal,we have:
$\angle ABC = \angle ACB$ .......... $(1)$
In $\Delta BDC$,we have
$BD = CD$ (Given,since $\Delta BDC$ is an isosceles triangle)
Since angles opposite to equal sides are equal,we have:
$\angle CBD = \angle BCD$ .......... $(2)$
Adding equations $(1)$ and $(2)$,we get:
$\angle ABC + \angle CBD = \angle ACB + \angle BCD$
$\Rightarrow \angle ABD = \angle ACD$
Hence proved.

(N/A) In $\Delta ABC$,since $AB = AC$,we have $\angle ABC = \angle ACB$ (angles opposite to equal sides). Let $\angle ABC = \angle ACB = x$. Then $\angle BAC = 180^\circ - 2x$.
In $\Delta ACD$,since $AD = AB$ and $AB = AC$,we have $AD = AC$. Thus,$\angle ADC = \angle ACD$ (angles opposite to equal sides). Let $\angle ADC = \angle ACD = y$.
Since $BA$ is produced to $D$,$BD$ is a straight line. Therefore,$\angle BAC + \angle CAD = 180^\circ$.
Since $\angle CAD = 180^\circ - 2y$ (from $\Delta ACD$),we have $(180^\circ - 2x) + (180^\circ - 2y) = 180^\circ$.
$360^\circ - 2(x + y) = 180^\circ$.
$2(x + y) = 180^\circ$.
$x + y = 90^\circ$.
Since $\angle BCD = \angle ACB + \angle ACD = x + y$,we conclude $\angle BCD = 90^\circ$.

(N/A) In $\triangle ABC$,we have:
$AB = AC$ [Given]
Since the sides are equal,their opposite angles are also equal.
$\Rightarrow \angle ACB = \angle ABC$
We know that the sum of all angles in a triangle is $180^o$.
$\angle A + \angle B + \angle C = 180^o$
$90^o + \angle B + \angle C = 180^o$ [Since $\angle A = 90^o$]
$\angle B + \angle C = 180^o - 90^o = 90^o$
Since $\angle B = \angle C$,we can write:
$2 \angle B = 90^o$
$\angle B = \frac{90^o}{2} = 45^o$
Therefore,$\angle B = 45^o$ and $\angle C = 45^o$.

(N/A) In $\Delta ABC$,we have:
$AB = BC = CA$ $[\because \Delta ABC$ is an equilateral triangle$]$
Since $AB = BC$,the angles opposite to these sides are equal,so $\angle A = \angle C$ ......... $(1)$
Similarly,since $AC = BC$,the angles opposite to these sides are equal,so $\angle A = \angle B$ ......... $(2)$
From $(1)$ and $(2)$,we have:
$\angle A = \angle B = \angle C$
Let $\angle A = \angle B = \angle C = x$
Since the sum of the angles in a triangle is $180^o$:
$\angle A + \angle B + \angle C = 180^o$
Substituting $x$ for each angle:
$x + x + x = 180^o$
$3x = 180^o$
$x = \frac{180^o}{3} = 60^o$
Therefore,$\angle A = \angle B = \angle C = 60^o$.
Thus,each angle of an equilateral triangle is $60^o$.

(N/A) In $\Delta DAC$,
$AD = AC$ (Given)
Therefore,$\angle ADC = \angle ACD$ (Angles opposite to equal sides are equal).
Now,$\angle ADC$ is an exterior angle for $\Delta ABD$.
By the exterior angle theorem,$\angle ADC > \angle ABD$.
Since $\angle ADC = \angle ACD$,we have $\angle ACD > \angle ABD$.
This implies $\angle ACB > \angle ABC$.
In $\Delta ABC$,the side opposite to the larger angle is longer. Therefore,$AB > AC$.
Since $AD = AC$,we conclude that $AB > AD$.

(N/A) Let us consider $\Delta ABC$ such that $\angle B = 90^\circ$.
Since the sum of angles in a triangle is $180^\circ$,we have $\angle A + \angle B + \angle C = 180^\circ$.
Substituting $\angle B = 90^\circ$,we get $\angle A + 90^\circ + \angle C = 180^\circ$,which implies $\angle A + \angle C = 90^\circ$.
Since $\angle A + \angle C = 90^\circ$ and $\angle B = 90^\circ$,it follows that $\angle B = \angle A + \angle C$.
This implies that $\angle B > \angle A$ and $\angle B > \angle C$.
In any triangle,the side opposite to the larger angle is longer. Therefore,the side opposite to $\angle B$ (which is $AC$) is longer than the side opposite to $\angle A$ (which is $BC$) and the side opposite to $\angle C$ (which is $AB$).
Thus,$AC > BC$ and $AC > AB$.
Since $AC$ is the hypotenuse,it is the longest side of the triangle.

(N/A) We know that $\angle ABC + \angle PBC = 180^\circ$ [Linear pair axiom].
Similarly,$\angle ACB + \angle QCB = 180^\circ$ [Linear pair axiom].
Therefore,$\angle ABC + \angle PBC = \angle ACB + \angle QCB$.
Given that $\angle PBC < \angle QCB$.
Since the sum of the angles is constant $(180^\circ)$,if one angle in the sum is smaller,the other must be larger to maintain the equality.
Thus,$\angle ABC > \angle ACB$.
In any triangle,the side opposite to the larger angle is longer.
Therefore,the side opposite to $\angle ABC$ (which is $AC$) is greater than the side opposite to $\angle ACB$ (which is $AB$).
Hence,$AC > AB$.

(N/A) Given: $\angle B < \angle A$ and $\angle C < \angle D$.
In $\triangle AOB$,since $\angle B < \angle A$,the side opposite to the greater angle is longer. Therefore,$OA < OB$ (side opposite to $\angle B$ is $OA$ and side opposite to $\angle A$ is $OB$) ...... $(1)$
In $\triangle COD$,since $\angle C < \angle D$,the side opposite to the greater angle is longer. Therefore,$OD < OC$ (side opposite to $\angle C$ is $OD$ and side opposite to $\angle D$ is $OC$) ...... $(2)$
Adding equations $(1)$ and $(2)$,we get:
$OA + OD < OB + OC$
Since $OA + OD = AD$ and $OB + OC = BC$,we have:
$AD < BC$.

(N/A) In $\Delta PQR$,$PS$ bisects $\angle QPR$ [Given].
Therefore,$\angle QPS = \angle RPS$.
Since $PR > PQ$ [Given],
Therefore,the angle opposite to $PR$ is greater than the angle opposite to $PQ$.
Thus,$\angle PQS > \angle PRS$.
Adding $\angle QPS$ to the left side and $\angle RPS$ to the right side (since $\angle QPS = \angle RPS$):
$\angle PQS + \angle QPS > \angle PRS + \angle RPS$ ... $(1)$.
We know that the exterior angle of a triangle is equal to the sum of the two interior opposite angles.
For $\Delta PQS$,the exterior angle $\angle PSR = \angle PQS + \angle QPS$.
For $\Delta PRS$,the exterior angle $\angle PSQ = \angle PRS + \angle RPS$.
Substituting these into $(1)$,we get $\angle PSR > \angle PSQ$.

(N/A) Let us consider a point $P$ not on line $l$. Let $PM$ be the perpendicular from $P$ to line $l$,and let $N$ be any other point on line $l$ such that $N \neq M$.
In $\Delta PMN$,since $\angle M = 90^o$,the sum of angles is $\angle M + \angle N + \angle P = 180^o$.
This implies $\angle N + \angle P = 90^o$,so $\angle N < 90^o$.
Since $\angle N < \angle M$,the side opposite to $\angle N$ must be smaller than the side opposite to $\angle M$.
Therefore,$PM < PN$.
Since $N$ is an arbitrary point on line $l$ (other than $M$),this shows that $PM$ is shorter than any other line segment drawn from $P$ to line $l$.
Thus,the perpendicular segment is the shortest line segment drawn from a point to a line.

(N/A) Let us consider a $\Delta ABC$.
Draw line $l$ as the perpendicular bisector of side $AB$.
Draw line $m$ as the perpendicular bisector of side $BC$.
Let the two perpendicular bisectors $l$ and $m$ intersect at point $O$.
Since any point on the perpendicular bisector of a segment is equidistant from its endpoints,point $O$ lies on the perpendicular bisector of $AB$ (so $OA = OB$) and on the perpendicular bisector of $BC$ (so $OB = OC$).
Therefore,$OA = OB = OC$,which means point $O$ is the required point equidistant from all vertices $A, B,$ and $C$. This point $O$ is known as the circumcenter of the triangle.

(N/A) Consider a $\Delta ABC$.
Draw the angle bisector $'l'$ of $\angle B$.
Draw the angle bisector $'m'$ of $\angle C$.
Let the two angle bisectors $'l'$ and $'m'$ intersect at point $O$.
Since any point on the angle bisector of an angle is equidistant from the arms of the angle,point $O$ lies on the bisector of $\angle B$ (so it is equidistant from $AB$ and $BC$) and also on the bisector of $\angle C$ (so it is equidistant from $BC$ and $AC$).
Therefore,point $O$ is equidistant from all three sides ($AB$,$BC$,and $AC$) of $\Delta ABC$. This point $O$ is known as the incenter of the triangle.

(N/A) To ensure the ice cream parlour is equidistant from points $A$,$B$,and $C$,we need to find the circumcenter of the triangle formed by these points.
$1$. Join points $A$ and $B$,and draw line $l$,which is the perpendicular bisector of segment $AB$.
$2$. Join points $B$ and $C$,and draw line $m$,which is the perpendicular bisector of segment $BC$.
$3$. Let the perpendicular bisectors $l$ and $m$ intersect at point $O$.
Point $O$ is the circumcenter of $\triangle ABC$,and it is equidistant from vertices $A$,$B$,and $C$. Therefore,the ice cream parlour should be set up at point $O$.

(D) In $\triangle ABC$,we are given that $AB = AC$.
Since the angles opposite to equal sides of a triangle are equal,we have $\angle C = \angle B$.
Given that $\angle B = 50^{\circ}$.
Therefore,$\angle C = 50^{\circ}$.

(A) In $\triangle ABC$,we are given that $BC = AB$.
Since the sides opposite to equal angles are equal,the angles opposite to equal sides are also equal.
Therefore,$\angle A = \angle C$.
We know that the sum of all interior angles of a triangle is $180^{\circ}$.
So,$\angle A + \angle B + \angle C = 180^{\circ}$.
Substituting the given values: $\angle A + 80^{\circ} + \angle A = 180^{\circ}$.
$2 \angle A + 80^{\circ} = 180^{\circ}$.
$2 \angle A = 180^{\circ} - 80^{\circ}$.
$2 \angle A = 100^{\circ}$.
$\angle A = 100^{\circ} / 2 = 50^{\circ}$.

(B) In $\triangle PQR$,we are given that $\angle R = \angle P$.
According to the property of triangles,the sides opposite to equal angles are equal.
Therefore,the side opposite to $\angle R$ (which is $PQ$) must be equal to the side opposite to $\angle P$ (which is $QR$).
Thus,$PQ = QR$.
Given that $QR = 4 \, cm$,it follows that $PQ = 4 \, cm$.

(C) In $\triangle ADC$,the exterior angle $\angle ADB$ is greater than the interior opposite angle $\angle DAC$.
Since $AD$ is the bisector of $\angle BAC$,we have $\angle BAD = \angle DAC$.
Substituting this into the inequality,we get $\angle ADB > \angle BAD$.
In $\triangle ABD$,since the side opposite to the greater angle is longer,we have $AB > BD$.

(A) According to the triangle inequality theorem,the sum of any two sides of a triangle must be greater than the third side,and the difference between any two sides must be less than the third side.
Let the third side be $x$.
Then,$5 - 1.5 < x < 5 + 1.5$,which simplifies to $3.5 < x < 6.5$.
Checking the options:
$A) 3.4$ is not in the range $(3.5, 6.5)$ because $3.4 < 3.5$.
$B) 3.8$ is in the range.
$C) 4.1$ is in the range.
$D) 3.6$ is in the range.
Therefore,the length of the third side cannot be $3.4\, cm$.

(B) In $\triangle PQR,$ we are given that $\angle R > \angle Q.$
According to the theorem,in any triangle,the side opposite to the greater angle is longer.
The side opposite to $\angle R$ is $PQ$ and the side opposite to $\angle Q$ is $PR.$
Since $\angle R > \angle Q,$ it follows that $PQ > PR.$

(N/A) According to the triangle inequality theorem,the sum of the lengths of any two sides of a triangle must be strictly greater than the length of the third side.
In this case,the lengths of the sides are $4 \, cm, 3 \, cm$,and $7 \, cm$.
Calculating the sum of the two smaller sides: $4 \, cm + 3 \, cm = 7 \, cm$.
Since the sum of the two sides $(7 \, cm)$ is equal to the third side $(7 \, cm)$ and not greater than it,the triangle inequality condition is not satisfied.
Therefore,it is not possible to construct a triangle with these side lengths.

(PR) In $\triangle PQR$,the sum of all interior angles is $180^{\circ}$.
$\angle Q = 180^{\circ} - (\angle P + \angle R)$
$\angle Q = 180^{\circ} - (70^{\circ} + 30^{\circ}) = 180^{\circ} - 100^{\circ} = 80^{\circ}$.
Comparing the angles,we have $\angle Q > \angle P > \angle R$ $(80^{\circ} > 70^{\circ} > 30^{\circ})$.
According to the property of triangles,the side opposite to the largest angle is the longest side.
The angle $\angle Q$ is the largest angle,and the side opposite to $\angle Q$ is $PR$.
Therefore,$PR$ is the longest side of $\triangle PQR$.

(A) We have to prove that $AB + BC + AC > 2\, AM$.
According to the triangle inequality theorem,the sum of any two sides of a triangle is greater than the third side.
In $\triangle ABM$,we have:
$AB + BM > AM$ $\ldots (1)$
In $\triangle ACM$,we have:
$AC + CM > AM$ $\ldots (2)$
Adding equations $(1)$ and $(2)$,we get:
$AB + BM + AC + CM > AM + AM$
$AB + AC + (BM + CM) > 2\, AM$
Since $M$ is a point on $BC$,$BM + CM = BC$. Substituting this into the inequality:
$AB + AC + BC > 2\, AM$
Thus,the perimeter of the triangle $ABC$ is indeed greater than $2\, AM$.

(N/A) No,it is not possible to construct a triangle with side lengths $9 \, cm$,$7 \, cm$,and $17 \, cm$.
According to the triangle inequality theorem,the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.
Here,the sum of the two smaller sides is:
$9 \, cm + 7 \, cm = 16 \, cm$
Since $16 \, cm < 17 \, cm$,the sum of the two sides is not greater than the third side.
Therefore,a triangle cannot be formed with these side lengths.

(A) Yes,it is possible to construct a triangle with side lengths $8\,cm, 7\,cm$ and $4\,cm$.
According to the triangle inequality theorem,the sum of any two sides of a triangle must be greater than the third side.
Checking the conditions:
$8 + 7 = 15 > 4$
$8 + 4 = 12 > 7$
$7 + 4 = 11 > 8$
Since all three conditions are satisfied,the triangle can be constructed.

(N/A) Given: In $\triangle PQR$,$PQ > PR$ and $QS$,$RS$ are bisectors of $\angle Q$ and $\angle R$ respectively.
Since $PQ > PR$,the angle opposite to the longer side is greater. Therefore,$\angle R > \angle Q$.
Since $QS$ and $RS$ are bisectors of $\angle Q$ and $\angle R$ respectively,we have:
$\angle SQR = \frac{1}{2} \angle Q$
$\angle SRQ = \frac{1}{2} \angle R$
Since $\angle R > \angle Q$,it follows that $\frac{1}{2} \angle R > \frac{1}{2} \angle Q$.
Therefore,$\angle SRQ > \angle SQR$.
In $\triangle SQR$,since $\angle SRQ > \angle SQR$,the side opposite to the greater angle is longer. Thus,$SQ > SR$.

(N/A) Given: $PQ = PR$.
To prove: $PS > PQ$.
Proof: In $\triangle PQR$,we have:
$PR = PQ$ (Given).
Therefore,$\angle PQR = \angle PRQ$ (Angles opposite to equal sides of a triangle are equal).
In $\triangle PQS$,$\angle PQR$ is an exterior angle.
We know that the exterior angle of a triangle is greater than each of the interior opposite angles.
Therefore,$\angle PQR > \angle S$.
Since $\angle PQR = \angle PRQ$,we have $\angle PRQ > \angle S$.
In $\triangle PSR$,since $\angle PRQ > \angle S$,the side opposite to $\angle PRQ$ must be greater than the side opposite to $\angle S$.
Therefore,$PS > PR$.
Since $PR = PQ$,we can substitute $PQ$ for $PR$.
Thus,$PS > PQ$.
Hence proved.

(N/A) Given: $A$ point $S$ on side $QR$ of $\triangle PQR$.
To prove: $PQ + QR + RP > 2 \, PS$.
Proof: In $\triangle PQS$,we have:
$PQ + QS > PS$ ..... $(1)$
[Since the sum of the lengths of any two sides of a triangle must be greater than the third side]
Now,in $\triangle PSR$,we have:
$RS + RP > PS$ ..... $(2)$
[Since the sum of the lengths of any two sides of a triangle must be greater than the third side]
Adding $(1)$ and $(2)$,we get:
$PQ + QS + RS + RP > 2 \, PS$
Since $QS + RS = QR$,we have:
$PQ + QR + RP > 2 \, PS$
Hence,proved.

(N/A) In $\triangle ABC$,we have $AB = AC$ (Given).
Therefore,$\angle ABC = \angle ACB$ (Angles opposite to equal sides of a triangle are equal).
Since $D$ is a point on $AC$,$\angle DBC < \angle ABC$.
Substituting $\angle ACB$ for $\angle ABC$,we get $\angle DBC < \angle ACB$.
Since $\angle ACB$ is the same as $\angle DCB$,we have $\angle DBC < \angle DCB$.
In $\triangle BCD$,since $\angle DCB > \angle DBC$,the side opposite to the greater angle is longer.
Therefore,$BD > CD$ or $CD < BD$.

(N/A) Given: In $\triangle ABC$,$AB = AC$. $BO$ and $CO$ are the angle bisectors of $\angle B$ and $\angle C$ respectively,intersecting at $O$. $BO$ is produced to $M$.
Proof:
In $\triangle ABC$,since $AB = AC$,we have $\angle ABC = \angle ACB$ (Angles opposite to equal sides are equal).
Since $BO$ and $CO$ are bisectors of $\angle B$ and $\angle C$,we have:
$\angle OBC = \frac{1}{2} \angle ABC$
$\angle OCB = \frac{1}{2} \angle ACB$
Since $\angle ABC = \angle ACB$,it follows that $\angle OBC = \angle OCB$.
In $\triangle OBC$,$\angle MOC$ is an exterior angle at vertex $O$ when side $BO$ is produced to $M$.
By the Exterior Angle Property of a triangle,the exterior angle is equal to the sum of the two interior opposite angles.
Therefore,$\angle MOC = \angle OBC + \angle OCB$.
Since $\angle OBC = \angle OCB$,we can write:
$\angle MOC = \angle OBC + \angle OBC = 2 \angle OBC$.
Substituting $\angle OBC = \frac{1}{2} \angle ABC$:
$\angle MOC = 2 \times (\frac{1}{2} \angle ABC) = \angle ABC$.
Hence,$\angle MOC = \angle ABC$.

(N/A) In $\triangle ABC$,we have $AB = AC$.
Therefore,$\angle ABC = \angle ACB$ [Since angles opposite to equal sides of a triangle are equal].
Since $BO$ and $CO$ are bisectors of $\angle B$ and $\angle C$ respectively,we have $\angle OBC = \frac{1}{2} \angle ABC$ and $\angle OCB = \frac{1}{2} \angle ACB$.
Since $\angle ABC = \angle ACB$,it follows that $\angle OBC = \angle OCB$.
In $\triangle OBC$,the sum of angles is $180^{\circ}$,so $\angle BOC + \angle OBC + \angle OCB = 180^{\circ}$.
Substituting $\angle OCB = \angle OBC$,we get $\angle BOC + 2 \angle OBC = 180^{\circ}$,which implies $\angle BOC = 180^{\circ} - 2 \angle OBC$.
Now,consider the external angle adjacent to $\angle ABC$. Let $D$ be a point on the extension of $AB$. The external angle is $\angle CBD = 180^{\circ} - \angle ABC$.
Since $\angle ABC = 2 \angle OBC$,we have $\angle CBD = 180^{\circ} - 2 \angle OBC$.
Comparing the two expressions,we conclude that $\angle BOC = \angle CBD$.

(N/A) Let $\angle BAD = \angle 1$ and $\angle CAD = \angle 2$. Since $AD$ is the bisector of $\angle BAC$,we have $\angle 1 = \angle 2$.
In $\triangle ACD$,$\angle ADC$ is an exterior angle at vertex $D$.
We know that an exterior angle of a triangle is equal to the sum of the two interior opposite angles.
Thus,$\angle ADC = \angle CAD + \angle ACD = \angle 2 + \angle ACD$.
Since $\angle ACD > 0$,it follows that $\angle ADC > \angle 2$.
Substituting $\angle 2 = \angle 1$,we get $\angle ADC > \angle 1$.
Now,consider $\triangle ABD$. In this triangle,the angle opposite to side $AB$ is $\angle ADB$ (which is $\angle ADC$) and the angle opposite to side $BD$ is $\angle BAD$ (which is $\angle 1$).
Since $\angle ADC > \angle 1$,the side opposite to the greater angle is longer.
Therefore,$AB > BD$.

(B) median of a triangle is a line segment joining a vertex to the midpoint of the opposite side. According to the theorem in geometry,triangles on the same base and between the same parallels are equal in area. Since the median divides the triangle into two triangles that share the same altitude and have bases of equal length (as the median bisects the side),the areas of these two triangles are equal.

(A) Let the side length of the equilateral triangle $ABC$ be $x$.
Since $ABC$ is an equilateral triangle,its area is given by $ar(\triangle ABC) = \frac{\sqrt{3}}{4} x^2$.
Given that $D$ is the mid-point of $BC$,the side length of the equilateral triangle $BDE$ is $BD = \frac{BC}{2} = \frac{x}{2}$.
The area of the equilateral triangle $BDE$ is $ar(\triangle BDE) = \frac{\sqrt{3}}{4} \left(\frac{x}{2}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{x^2}{4} = \frac{1}{4} \left(\frac{\sqrt{3}}{4} x^2\right)$.
Substituting the area of $\triangle ABC$,we get $ar(\triangle BDE) = \frac{1}{4} ar(\triangle ABC)$.
Therefore,the given statement is True.

(A) Let the side length of the equilateral triangle $ABC$ be $a$.
Since $D$ is the mid-point of $BC$,the side length of the equilateral triangle $BDE$ is $\frac{a}{2}$.
The area of an equilateral triangle with side $s$ is given by $\frac{\sqrt{3}}{4} s^2$.
$\operatorname{ar}(\triangle ABC) = \frac{\sqrt{3}}{4} a^2$.
$\operatorname{ar}(\triangle BDE) = \frac{\sqrt{3}}{4} \left(\frac{a}{2}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{a^2}{4} = \frac{1}{4} \left(\frac{\sqrt{3}}{4} a^2\right)$.
Therefore,$\operatorname{ar}(\triangle BDE) = \frac{1}{4} \operatorname{ar}(\triangle ABC)$.
Hence,the given statement is True.

(B) The given statement is False.
In any triangle $ABC$,the sum of all interior angles must be equal to $180^{\circ}$,i.e.,$\angle A + \angle B + \angle C = 180^{\circ}$.
Given that $\angle B = 105^{\circ}$ and $\angle C = 90^{\circ}$.
Calculating the sum of the two given angles: $\angle B + \angle C = 105^{\circ} + 90^{\circ} = 195^{\circ}$.
Since $195^{\circ} > 180^{\circ}$,the sum of two angles alone exceeds the angle sum property of a triangle. Therefore,such a triangle cannot be constructed.

(N/A) Produce $CB$ to a point $D$ such that $BC = BD$ and join $AD$.
In $\Delta ABC$ and $\Delta ABD$,we have:
$BC = BD$ (By construction)
$AB = AB$ (Common side)
$\angle ABC = \angle ABD = 90^{\circ}$ (Given and by construction)
Therefore,$\Delta ABC \cong \Delta ABD$ by $SAS$ congruence rule.
So,$\angle CAB = \angle DAB$ by $CPCT$. Let $\angle BAC = x$,then $\angle CAB = x$ and $\angle DAB = x$.
Thus,$\angle CAD = \angle CAB + \angle BAD = x + x = 2x$.
Since $\angle BCA = 2 \angle BAC$,and $\angle BAC = x$,then $\angle BCA = 2x$.
In $\Delta ACD$,$\angle CAD = 2x$ and $\angle ACD = 2x$.
Since two angles are equal,the sides opposite to them are equal,so $AC = CD$.
Also,since $\Delta ABC \cong \Delta ABD$,$AC = AD$ by $CPCT$.
Thus,$\angle ADC = \angle ACD = 2x$.
Since all three angles of $\Delta ACD$ are $60^{\circ}$ (as $2x + 2x + 2x = 180^{\circ} \implies x = 30^{\circ}$),$\Delta ACD$ is an equilateral triangle.
Therefore,$AC = CD = AD$.
Since $CD = BC + BD$ and $BC = BD$,we have $CD = BC + BC = 2BC$.
Hence,$AC = 2BC$.

(N/A) We are given a point $D$ on side $BC$ of a $\Delta ABC$ such that $\angle BAD = \angle CAD$ and $BD = CD$. We are to prove that $AB = AC$.
Produce $AD$ to a point $E$ such that $AD = DE$ and then join $CE$.
Now,in $\Delta ABD$ and $\Delta ECD$,we have:
$BD = CD$ (Given)
$AD = DE$ (By construction)
$\angle ADB = \angle EDC$ (Vertically opposite angles)
Therefore,$\Delta ABD \cong \Delta ECD$ ($SAS$ congruence rule).
So,$AB = EC$ and $\angle BAD = \angle CED$ $(CPCT)$ ... $(1)$
Also,$\angle BAD = \angle CAD$ (Given) ... $(2)$
From $(1)$ and $(2)$,we get $\angle CAD = \angle CED$.
In $\Delta ACE$,since $\angle CAD = \angle CED$,the sides opposite to these angles must be equal.
Therefore,$AC = EC$ ... $(3)$
From $(1)$ and $(3)$,we have $AB = AC$.
Hence,$\Delta ABC$ is an isosceles triangle.