Parallel Lines and a Transversal Questions in English

(B) To solve this,we draw a line $AB$ passing through point $M$ such that $AB \parallel PQ$. Since $PQ \parallel RS$ and $AB \parallel PQ$,it follows that $AB \parallel RS$.
Now,consider the transversal $XM$ intersecting parallel lines $PQ$ and $AB$. The interior angles on the same side of the transversal are supplementary.
Therefore,$\angle QXM + \angle XMB = 180^o$.
Given $\angle QXM = 135^o$,we have $135^o + \angle XMB = 180^o$,which gives $\angle XMB = 180^o - 135^o = 45^o$ (Equation $1$).
Next,consider the transversal $YM$ intersecting parallel lines $RS$ and $AB$. The alternate interior angles are equal.
Therefore,$\angle BMY = \angle MYR$.
Given $\angle MYR = 40^o$,we have $\angle BMY = 40^o$ (Equation $2$).
Finally,$\angle XMY = \angle XMB + \angle BMY$.
Substituting the values from Equation $1$ and Equation $2$,we get $\angle XMY = 45^o + 40^o = 85^o$.

(N/A) In the figure,a transversal $AD$ intersects two lines $PQ$ and $RS$ at points $B$ and $C$ respectively. Ray $BE$ is the bisector of $\angle ABQ$ and ray $CG$ is the bisector of $\angle BCS$; and $BE \parallel CG$.
We are to prove that $PQ \parallel RS$.
It is given that ray $BE$ is the bisector of $\angle ABQ$.
Therefore,$\angle ABE = \frac{1}{2} \angle ABQ$ ...... $(1)$
Similarly,ray $CG$ is the bisector of $\angle BCS$.
Therefore,$\angle BCG = \frac{1}{2} \angle BCS$ ...... $(2)$
But $BE \parallel CG$ and $AD$ is the transversal.
Therefore,$\angle ABE = \angle BCG$ (Corresponding angles axiom) ...... $(3)$
Substituting $(1)$ and $(2)$ in $(3)$,you get
$\frac{1}{2} \angle ABQ = \frac{1}{2} \angle BCS$
That is,$\angle ABQ = \angle BCS$
But,they are the corresponding angles formed by transversal $AD$ with $PQ$ and $RS$; and are equal.
Therefore,$PQ \parallel RS$ (Converse of corresponding angles axiom).

(D) Given that $AB \parallel CD$ and $CD \parallel EF$,therefore $AB \parallel EF$.
Since $CD \parallel EF$ and $ED$ is a transversal,the interior angles on the same side of the transversal are supplementary.
Thus,$y + 55^o = 180^o$.
$y = 180^o - 55^o = 125^o$.
Since $AB \parallel CD$ and $BE$ is a transversal,the corresponding angles are equal.
Thus,$x = y = 125^o$.
Since $EA \perp AB$ and $AB \parallel EF$,then $EA \perp EF$. Therefore,$\angle AEF = 90^o$.
From the figure,$\angle AEF = z + 55^o$.
$z + 55^o = 90^o$.
$z = 90^o - 55^o = 35^o$.
Therefore,$x = 125^o, y = 125^o, z = 35^o$.

(N/A) From the figure,the line $PQ$ intersects the line $CD$ at point $F$.
Therefore,$y = 130^o$ (Vertically opposite angles) .......... $(1)$
Now,consider the line $AB$. The angle $x$ and the angle $50^o$ form a linear pair at the intersection point $E$.
Therefore,$x + 50^o = 180^o$ (Linear pair axiom)
$x = 180^o - 50^o = 130^o$ .......... $(2)$
From $(1)$ and $(2)$,we have $x = y = 130^o$.
Since these are alternate interior angles and they are equal,the lines $AB$ and $CD$ must be parallel.
Therefore,$AB \parallel CD$.

(B) Given: $AB \parallel CD$ and $CD \parallel EF$.
Since lines parallel to the same line are parallel to each other,we have $AB \parallel EF$.
Let the transversal intersect the lines $AB$,$CD$,and $EF$ at points $P$,$Q$,and $R$ respectively.
Since $AB \parallel EF$,the interior alternate angles are equal. Therefore,$\angle x = \angle z$ .......... $(1)$
Again,$AB \parallel CD$,so the sum of interior angles on the same side of the transversal is $180^o$.
Thus,$y + z = 180^o$.
Given $y: z = 3: 7$,let $y = 3k$ and $z = 7k$.
Then,$3k + 7k = 180^o \Rightarrow 10k = 180^o \Rightarrow k = 18^o$.
So,$z = 7 \times 18^o = 126^o$.
From $(1)$,$x = z = 126^o$.

(N/A) Given that $AB \parallel CD$,$EF \perp CD$,and $\angle GED = 126^o$.
$1$. Since $AB \parallel CD$ and $GE$ is a transversal,the alternate interior angles are equal.
Therefore,$\angle AGE = \angle GED$.
Given $\angle GED = 126^o$,so $\angle AGE = 126^o$.
$2$. We know that $\angle GED = \angle GEF + \angle FED$.
Since $EF \perp CD$,$\angle FED = 90^o$.
Therefore,$126^o = \angle GEF + 90^o$.
$\angle GEF = 126^o - 90^o = 36^o$.
$3$. Since $AB \parallel CD$ and $GE$ is a transversal,the sum of interior angles on the same side of the transversal is $180^o$.
Therefore,$\angle FGE + \angle GED = 180^o$.
$\angle FGE + 126^o = 180^o$.
$\angle FGE = 180^o - 126^o = 54^o$.
Thus,$\angle AGE = 126^o$,$\angle GEF = 36^o$,and $\angle FGE = 54^o$.

(D) Given: $PQ \parallel ST$,$\angle PQR = 110^o$,and $\angle RST = 130^o$.
Construction: Draw a line $EF$ parallel to $ST$ passing through point $R$.
Since $PQ \parallel ST$ and $EF \parallel ST$,it follows that $PQ \parallel EF$.
Since $PQ \parallel EF$ and $QR$ is a transversal,the sum of interior angles on the same side of the transversal is $180^o$ (or by alternate interior angles if extended). Alternatively,using alternate interior angles: $\angle PQR + \angle QRF = 180^o$ is incorrect here; rather,extend $PQ$ to meet $EF$ or use the property that $\angle PQR + \angle QRF = 180^o$ is not applicable directly. Let's use: $\angle PQR + \angle QRF = 180^o$ (consecutive interior angles) is wrong. Correct approach: $\angle PQR + \angle QRE = 180^o$ (consecutive interior angles) $\Rightarrow 110^o + \angle QRE = 180^o \Rightarrow \angle QRE = 70^o$.
Similarly,for $ST \parallel EF$ and transversal $RS$,$\angle RST + \angle SRE = 180^o$ (consecutive interior angles) $\Rightarrow 130^o + \angle SRE = 180^o \Rightarrow \angle SRE = 50^o$.
Now,$\angle QRS = \angle QRE + \angle SRE = 70^o + 50^o = 120^o$ is incorrect based on the figure. Let's re-evaluate: $\angle PQR = 110^o$. Draw $EF \parallel PQ$. Then $\angle PQR + \angle QRE = 180^o \Rightarrow \angle QRE = 70^o$. Since $PQ \parallel ST$ and $EF \parallel PQ$,then $EF \parallel ST$. Thus $\angle RST + \angle SRE = 180^o \Rightarrow 130^o + \angle SRE = 180^o \Rightarrow \angle SRE = 50^o$. From the figure,$\angle QRS = 360^o - (\angle QRE + \angle SRE)$ is not correct. Actually,$\angle QRS = 180^o - (70^o + 50^o) = 60^o$.

(A) Given: $AB \parallel CD$,$\angle APQ = 50^o$,and $\angle PRD = 127^o$.
$1$. Since $AB \parallel CD$ and $PQ$ is a transversal,the alternate interior angles are equal.
Therefore,$\angle PQR = \angle APQ$.
Given $\angle APQ = 50^o$,so $x = 50^o$.
$2$. Since $AB \parallel CD$ and $PR$ is a transversal,the alternate interior angles are equal.
Therefore,$\angle APR = \angle PRD$.
Given $\angle PRD = 127^o$,so $\angle APR = 127^o$.
We know that $\angle APR = \angle APQ + \angle QPR$.
Substituting the values,$127^o = 50^o + y$.
$y = 127^o - 50^o = 77^o$.
Thus,$x = 50^o$ and $y = 77^o$.

(N/A) $(i)$ Perpendiculars to parallel lines are parallel.
$(ii)$ According to the laws of reflection,the angle of incidence $=$ the angle of reflection.
Draw ray $BL \perp PQ$ and $CM \perp RS$.
$\because PQ \parallel RS$ [Given]
$\therefore BL \parallel CM$ [$\because BL \perp PQ$ and $CM \perp RS$]
And $BC$ is a transversal.
$\therefore \angle LBC = \angle MCB$ [Interior alternate angles] ......... $(1)$
Since,(Angle of incidence) $=$ (Angle of reflection)
$\therefore \angle ABL = \angle LBC$ and $\angle MCB = \angle MCD$
$\Rightarrow \angle ABL = \angle MCD$
$\therefore$ From $(1)$,we have
$\angle LBC + \angle ABL = \angle MCB + \angle MCD$ [Equals are added to equals]
$\Rightarrow \angle ABC = \angle BCD$
i.e.,the angles of a pair of interior alternate angles are equal.
$\therefore AB \parallel CD$.

(C) Given that $AB \parallel DE$ and $AE$ acts as a transversal.
Since $AB \parallel DE$,the alternate interior angles are equal.
Therefore,$\angle AED = \angle BAC = 35^o$.
Now,consider the triangle $\Delta CDE$.
The sum of the angles in a triangle is $180^o$.
So,$\angle CDE + \angle DEC + \angle DCE = 180^o$.
Substituting the known values,we get $53^o + 35^o + \angle DCE = 180^o$.
$88^o + \angle DCE = 180^o$.
$\angle DCE = 180^o - 88^o = 92^o$.

(A) When a transversal intersects two parallel lines,the sum of the interior angles on the same side of the transversal is $180^{\circ}$ (consecutive interior angles are supplementary).
Let the two angles be $2x$ and $3x$.
According to the property,$2x + 3x = 180^{\circ}$.
$5x = 180^{\circ}$.
$x = 180^{\circ} / 5 = 36^{\circ}$.
The two angles are $2(36^{\circ}) = 72^{\circ}$ and $3(36^{\circ}) = 108^{\circ}$.
The greater of the two angles is $108^{\circ}$.

(D) To solve this,draw a line $XY$ parallel to $OP$ passing through point $R.$ Since $OP \parallel RS,$ it follows that $XY \parallel OP \parallel RS.$
Now,consider the parallel lines $OP$ and $XY.$ Since $PQ$ is a transversal,the sum of consecutive interior angles is $180^{\circ}.$
$\angle OPQ + \angle PQR_{1} = 180^{\circ}$
$110^{\circ} + \angle PQR_{1} = 180^{\circ}$
$\angle PQR_{1} = 180^{\circ} - 110^{\circ} = 70^{\circ}$
Next,consider the parallel lines $XY$ and $RS.$ Since $QR$ is a transversal,the sum of consecutive interior angles is $180^{\circ}.$
$\angle QRS + \angle RQR_{2} = 180^{\circ}$
$130^{\circ} + \angle RQR_{2} = 180^{\circ}$
$\angle RQR_{2} = 180^{\circ} - 130^{\circ} = 50^{\circ}$
Finally,$\angle PQR = \angle PQR_{1} + \angle RQR_{2} = 70^{\circ} + 50^{\circ} = 120^{\circ}.$
Wait,re-evaluating the geometry: If we extend $OP$ to intersect $QR$ at $X,$ then $\angle RXP = 180^{\circ} - 130^{\circ} = 50^{\circ}$ (interior angles on same side of transversal). In $\triangle PXQ,$ $\angle OPQ$ is an exterior angle,so $\angle OPQ = \angle PXQ + \angle PQR.$
$110^{\circ} = 50^{\circ} + \angle PQR$
$\angle PQR = 110^{\circ} - 50^{\circ} = 60^{\circ}.$

(NO) No,the two lines will not always be parallel. For two lines to be parallel,the sum of the interior angles on the same side of the transversal must be $180^{\circ}$. If each of the two equal interior angles is $x$,then their sum is $x + x = 2x$. The lines are parallel only if $2x = 180^{\circ}$,which means $x = 90^{\circ}$. If $x \neq 90^{\circ}$,the lines are not parallel.

$(136^{\circ})$ If a transversal intersects two parallel lines,then each pair of consecutive interior angles (also known as co-interior angles) are supplementary,meaning their sum is $180^{\circ}$.
In the given figure,the angles $x$ and $44^{\circ}$ are consecutive interior angles on the same side of the transversal $n$.
Therefore,we have the equation:
$x + 44^{\circ} = 180^{\circ}$
Solving for $x$:
$x = 180^{\circ} - 44^{\circ}$
$x = 136^{\circ}$

(N/A) For Fig. $(i)$,a transversal intersects two lines such that the sum of interior angles on the same side of the transversal is $132^{\circ} + 48^{\circ} = 180^{\circ}$.
Since the sum of consecutive interior angles is $180^{\circ}$,the lines $l$ and $m$ are parallel.
For Fig. (ii),a transversal intersects two lines such that the sum of interior angles on the same side of the transversal is $73^{\circ} + 106^{\circ} = 179^{\circ}$.
Since the sum of consecutive interior angles is not $180^{\circ}$,the lines $p$ and $q$ are not parallel.

(B) When two lines $l$ and $m$ are perpendicular to the same line $n$,the interior angles on the same side of the transversal $n$ are both $90^{\circ}$.
The sum of these interior angles is $90^{\circ} + 90^{\circ} = 180^{\circ}$.
According to the property of parallel lines,if the sum of interior angles on the same side of a transversal is $180^{\circ}$,then the lines are parallel.
Therefore,the lines $l$ and $m$ are parallel to each other,not perpendicular.

(N/A) Given: $x=y$ and $a=b$.
$1$. For lines $l$ and $m$ with a transversal,the corresponding angles $x$ and $y$ are equal $(x=y)$. By the converse of the corresponding angles axiom,if corresponding angles are equal,then the lines are parallel. Therefore,$l \parallel m$ $....(1)$
$2$. For lines $n$ and $m$ with a transversal,the corresponding angles $a$ and $b$ are equal $(a=b)$. By the converse of the corresponding angles axiom,if corresponding angles are equal,then the lines are parallel. Therefore,$n \parallel m$ $....(2)$
$3$. From $(1)$ and $(2)$,since both lines $l$ and $n$ are parallel to the same line $m$,it follows that $l \parallel n$ (Lines parallel to the same line are parallel to each other).

(N/A) We are given that $\angle 1 = 60^{\circ}$ and $\angle 6 = 120^{\circ}$.
From the figure,$\angle 5$ and $\angle 6$ form a linear pair.
Therefore,$\angle 5 + \angle 6 = 180^{\circ}$.
Substituting the value of $\angle 6$,we get $\angle 5 + 120^{\circ} = 180^{\circ}$.
$\angle 5 = 180^{\circ} - 120^{\circ} = 60^{\circ}$.
Now,we observe that $\angle 1 = 60^{\circ}$ and $\angle 5 = 60^{\circ}$.
Thus,$\angle 1 = \angle 5$.
Since $\angle 1$ and $\angle 5$ are corresponding angles and they are equal,the lines $m$ and $n$ must be parallel.

(N/A) Given: $l \parallel m$ and $t$ is a transversal intersecting $l$ at $A$ and $m$ at $B$.
$AP$ is the bisector of $\angle MAB$ and $BQ$ is the bisector of $\angle SBA$ (where $S$ is a point on line $m$ such that $\angle MAB$ and $\angle SBA$ are alternate interior angles).
Since $l \parallel m$ and $t$ is a transversal,the alternate interior angles are equal:
$\angle MAB = \angle SBA$
Since $AP$ and $BQ$ are bisectors:
$\angle PAB = \frac{1}{2} \angle MAB$ and $\angle QBA = \frac{1}{2} \angle SBA$
Therefore,$\angle PAB = \angle QBA$.
These are alternate interior angles formed by the transversal $t$ with lines $AP$ and $BQ$.
Since the alternate interior angles are equal,the lines must be parallel.
Hence,$AP \parallel BQ$.

(N/A) Produce $DE$ to intersect $BC$ at $P$.
Since $EF \parallel BC$ and $DP$ is the transversal,
$\therefore \angle DEF = \angle DPC$ ....$(1)$ [Corresponding angles]
Now,$AB \parallel ED$ (which is the same line as $DP$),and $BC$ is the transversal,
$\therefore \angle DPC = \angle ABC$ ....$(2)$ [Corresponding angles]
From $(1)$ and $(2)$,we get
$\angle ABC = \angle DEF$
Hence,proved.

(N/A) Produce $ED$ to meet $BC$ at point $P$.
Since $EF \parallel BC$ and $EP$ is the transversal,the interior angles on the same side of the transversal are supplementary.
Therefore,$\angle DEF + \angle EPC = 180^{\circ} .....(1)$
Again,since $AB \parallel EP$ and $BC$ is the transversal,the corresponding angles are equal.
Therefore,$\angle ABC = \angle EPC .....(2)$
Substituting the value of $\angle EPC$ from equation $(2)$ into equation $(1)$,we get:
$\angle DEF + \angle ABC = 180^{\circ}$
Hence,$\angle ABC + \angle DEF = 180^{\circ}.$
Thus,it is proved.

(N/A) Let the two parallel lines be $l$ and $m$,such that $l \parallel m$. Let line $p$ be perpendicular to $l$ $(p \perp l)$ and line $n$ be perpendicular to $m$ $(n \perp m)$.
We need to show that $p \parallel n$.
Since $n \perp m$,the angle between them is $90^{\circ}$,so $\angle 1 = 90^{\circ}$.
Since $p \perp l$,the angle between them is $90^{\circ}$,so $\angle 2 = 90^{\circ}$.
Since $l \parallel m$ and $p$ acts as a transversal,the corresponding angles are equal,so $\angle 2 = \angle 3$.
Since $\angle 2 = 90^{\circ}$,it follows that $\angle 3 = 90^{\circ}$.
Now,considering lines $p$ and $n$ with transversal $m$,we have $\angle 1 = 90^{\circ}$ and $\angle 3 = 90^{\circ}$.
Since $\angle 1 = \angle 3$,these are corresponding angles,which implies that the lines $p$ and $n$ must be parallel.
Therefore,$p \parallel n$.

(N/A) Given: Two parallel lines $DE$ and $QR$ are intersected by a transversal at points $A$ and $B$ respectively. Let $AF$ be the bisector of $\angle CAE$ and $BP$ be the bisector of $\angle ABR$.
To prove: $AF \parallel BP$.
Proof: Since $DE \parallel QR$,the corresponding angles are equal,so $\angle CAE = \angle ABR$.
Multiplying both sides by $\frac{1}{2}$,we get $\frac{1}{2} \angle CAE = \frac{1}{2} \angle ABR$.
Since $AF$ and $BP$ are bisectors,$\frac{1}{2} \angle CAE = \angle FAB$ and $\frac{1}{2} \angle ABR = \angle ABP$.
Therefore,$\angle FAB = \angle ABP$.
These are corresponding angles formed by the transversal $n$ intersecting lines $AF$ and $BP$. Since the corresponding angles are equal,the lines must be parallel. Hence,$AF \parallel BP$.

(C) $1$. From the figure,the angle $x$ and the angle $70^{\circ}$ are vertically opposite angles. Therefore,$x = 70^{\circ}$.
$2$. The angle $y$ and the angle $110^{\circ}$ form a linear pair on the line $PQ$. Therefore,$y + 110^{\circ} = 180^{\circ}$,which gives $y = 180^{\circ} - 110^{\circ} = 70^{\circ}$. Wait,looking at the figure,$y$ is vertically opposite to $110^{\circ}$,so $y = 110^{\circ}$.
$3$. Alternatively,the interior angles on the same side of the transversal are $y$ and $x$. Since $y = 110^{\circ}$ and $x = 70^{\circ}$,their sum is $110^{\circ} + 70^{\circ} = 180^{\circ}$.
$4$. Since the sum of the interior angles on the same side of the transversal is $180^{\circ}$,the lines $PQ$ and $XY$ must be parallel $(PQ \parallel XY)$.

(N/A) Given: $AB \parallel DE$ and $BC \parallel EF$.
To prove: $\angle ABC = \angle DEF$.
Construction: Extend $DE$ to intersect $BC$ at point $G$.
Proof:
$1$. Since $AB \parallel DE$ and $BC$ is a transversal,$\angle ABC = \angle DGC$ (Corresponding angles).
$2$. Since $BC \parallel EF$ and $DE$ is a transversal,$\angle DGC = \angle DEF$ (Corresponding angles).
$3$. From equations $(1)$ and $(2)$,we get $\angle ABC = \angle DEF$.
Hence proved.

(N/A) Given: $AB \parallel DE$ and $BC \parallel EF$.
To prove: $\angle ABC + \angle DEF = 180^{\circ}$.
Construction: Extend $ED$ to meet $BC$ at point $G$.
Proof:
Since $AB \parallel DE$ and $ED$ is extended to $G$,we have $AB \parallel DG$.
Since $AB \parallel DG$ and $BG$ is a transversal,the interior angles on the same side are supplementary,but here $AB$ and $DG$ are parallel and $BC$ is a transversal,so $\angle ABC + \angle BGD = 180^{\circ}$ (Consecutive interior angles).
Now,consider $BC \parallel EF$. Since $BC \parallel EF$ and $EG$ is a transversal,$\angle BGD = \angle DEF$ (Corresponding angles).
Substituting $\angle BGD = \angle DEF$ in the first equation,we get $\angle ABC + \angle DEF = 180^{\circ}$.
Hence proved.

(B) When two parallel lines are intersected by a transversal,the interior angles on the same side of the transversal are consecutive interior angles (also known as co-interior angles).
According to the properties of parallel lines,these angles are supplementary,meaning their sum is equal to $180^{\circ}$.

(A) When two parallel lines are intersected by a transversal,the corresponding angles formed at each intersection are equal. This is a fundamental property of parallel lines known as the Corresponding Angles Axiom.

(N/A) Let the two parallel lines be $l$ and $m$,and let $n$ be the transversal intersecting them at points $A$ and $B$ respectively.
Let the interior angles formed be $\angle PAB$,$\angle QAB$,$\angle RBA$,and $\angle SBA$.
The bisectors of these four interior angles form a quadrilateral $EFGH$.
Since $l \parallel m$,the sum of consecutive interior angles is $180^{\circ}$,i.e.,$\angle PAB + \angle RBA = 180^{\circ}$.
Dividing by $2$,we get $\frac{1}{2}\angle PAB + \frac{1}{2}\angle RBA = 90^{\circ}$.
In $\triangle EAB$,$\angle EAB + \angle EBA = 90^{\circ}$,which implies $\angle AEB = 90^{\circ}$.
Similarly,all other angles of the quadrilateral $EFGH$ can be shown to be $90^{\circ}$.
Since all interior angles are $90^{\circ}$,the quadrilateral $EFGH$ is a rectangle.