Solution

As $\mathrm{\angle }ARQ=\mathrm{\angle }RQD={25}^{\circ }$ [alt. $\mathrm{\angle }s$ ]

Also, $\mathrm{\angle }RQC={180}^{\circ }-{60}^{\circ }={120}^{\circ }$ (linear pair)

And, $\mathrm{\angle }SRA={120}^{\circ }$ (Corresponding angle)

Now,

$\mathrm{\angle }SRQ={120}^{\circ }+{25}^{\circ }$

$\mathrm{\angle }SRQ={145}^{\circ }$

Hence, the correct option is (C).

Solution

Let angle of triangle $ABC$ be $\mathrm{\angle }A,\mathrm{\angle }B$ and $\mathrm{\angle }C$

Given that:

$\mathrm{\angle }A=\mathrm{\angle }B+\mathrm{\angle }C$

We know that in any triangle $\mathrm{\angle }A+\mathrm{\angle }B+\mathrm{\angle }C={180}^{\circ }$

From equation (I) and (II), get:

$\mathrm{\angle }A+\mathrm{\angle }A={180}^{\circ }$

$\begin{array}{rl}2\mathrm{\angle }A& ={180}^{\circ }\\ \mathrm{\angle }A& ={\displaystyle \frac{{180}^{\circ }}{2}}\\ \mathrm{\angle }A& ={90}^{\circ }\end{array}$

Hence, the triangle is a right triangle.

Therefore, the correct option is (D).

Solution

Given: An exterior angle of triangle is ${150}^{\circ }$.

Let each of the two interior opposite angle be $x$.

The sum of two interior opposite angle is equal to exterior angle of a triangle.

$\begin{array}{rl}{105}^{\circ }& =x+x\\ 2x& ={105}^{\circ }\\ x& =52{{\displaystyle \frac{1}{2}}}^{\circ }\end{array}$

So, each of equal angle is $52{{\displaystyle \frac{1}{2}}}^{\circ }$

Hence, the correct option is (B).

Solution

Let the angle of the triangle are $5x,3x$ and $7x$. As we know that sum of all angle of triangle is ${180}^{\circ }$. Now,

$5x+3x+7x={180}^{\circ }$

$\begin{array}{rl}15x& ={180}^{\circ }\\ x& ={\displaystyle \frac{{180}^{\circ }}{15}}\\ x& ={12}^{\circ }\end{array}$

Hence, the angle of the triangle are:

$5\times {12}^{\circ }={60}^{\circ }$

$3\times {12}^{\circ }={36}^{\circ }$

$7\times {12}^{\circ }={84}^{\circ }$

All the angle of this triangle is less than 90 degree.

Hence, the triangle is an acute angled triangle.

Solution

In triangle $ABC$, Let $\mathrm{\angle }A={130}^{\circ }$.

The bisector of the angle $B$ and $C$ are $OB$ and $OC$.

Let $\mathrm{\angle }OBC=\mathrm{\angle }OBA=x$ and $\mathrm{\angle }OCB=\mathrm{\angle }OCA=y$

In triangle $ABC$,

$\begin{array}{rl}\mathrm{\angle }A+\mathrm{\angle }B+\mathrm{\angle }C& ={180}^{\circ }\\ {130}^{\circ }+2x+2y& ={180}^{\circ }\\ 2x+2y& ={180}^{\circ }-{130}^{\circ }\\ 2x+2y& ={50}^{\circ }\\ x+y& ={25}^{\circ }\end{array}$

That is $\mathrm{\angle }OBC+\mathrm{\angle }OCA={25}^{\circ }$

Now, in triangle BOC:

$\begin{array}{rl}\mathrm{\angle }BOC& ={180}^{\circ }-(\mathrm{\angle }OBC+\mathrm{\angle }OCB)\\ & ={180}^{\circ }-{25}^{\circ }\\ & ={155}^{\circ }\end{array}$

Hence, the correct option is (D).

Solution

See the given figure in the question:

$\begin{array}{c}{40}^{\circ }+4x+3x={180}^{\circ }\text{ (Angles on the straight line) }\\ 4x+3x={180}^{\circ }-{40}^{\circ }\\ 7x={140}^{\circ }\\ x={\displaystyle \frac{{140}^{\circ }}{7}}\\ x={20}^{\circ }\end{array}$

Hence, the correct option is (A).

Solution

See the given figure, producing OP, to intersect RQ at X.

Given: $OP||RS$ and $RX$ is a transversal.

So, $\mathrm{\angle }RXP=\mathrm{\angle }XRS$ (alternative angle)

$\mathrm{\angle }RXP={130}^{\circ }$ [Given: $\mathrm{\angle }QRS={130}^{\circ }$ ]

$RQ$ is a line segment.

So, $\mathrm{\angle }PXQ+\mathrm{\angle }RXV={180}^{\circ }$ [linear pair axiom]

$\mathrm{\angle }PXQ={180}^{\circ }-\mathrm{\angle }RXP={180}^{\circ }-{130}^{\circ }$

$\mathrm{\angle }PXQ={50}^{\circ }$

In triangle $PQX,\mathrm{\angle }OPQ$ is an exterior angle,

Therefore, $\mathrm{\angle }OPQ=\mathrm{\angle }PXQ+\mathrm{\angle }PQX$ [exterior angle = sum of two opposite interior angles]

${110}^{\circ }={50}^{\circ }+\mathrm{\angle }PQX$

$\mathrm{\angle }PQX={110}^{\circ }-{50}^{\circ }$

$\mathrm{\angle }PQR={60}^{\circ }$

Hence, the correct option is (

Solution

Given, the ratio of angles of a triangle is $2:4:3$.

Let the angles of a triangle be $\mathrm{\angle }A,\mathrm{\angle }B$ and $\mathrm{\angle }C$.

$\mathrm{\angle }A=2x,\mathrm{\angle }B=4x\mathrm{\angle }C=3x$,

$\mathrm{\angle }A+\mathrm{\angle }B+\mathrm{\angle }C={180}^{\circ }$ [sum of all the angles of a triangle is ${180}^{\circ }$ ]

$2x+4x+3x={180}^{\circ }$

$9x={180}^{\circ }$

$x={180}^{\circ }/9$

$={20}^{\circ }$

$\mathrm{\angle }A=2x=2\times {20}^{\circ }={40}^{\circ }$

$\mathrm{\angle }B=4x=4x{20}^{\circ }={80}^{\circ }$

$\mathrm{\angle }C=3x=3\times {20}^{\circ }={60}^{\circ }$

So, the smallest angle of a triangle is ${40}^{\circ }$.

Hence, the correct option is (B).

Solution

See the figure, $x$ and $y$ are two adjacent angles.

For $ABC$ to be a straight line, the sum of two adjacent angle must be ${180}^{\circ }$.

Solution

We know that in a triangle, sum of all the angles is always ${180}^{\circ }$. So, a triangle can’t have all angles less than ${60}^{\circ }$.

Solution

If an angle whose measure is more than ${90}^{\circ }$ but less than ${180}^{\circ }$ is called an obtuse angle.

We know that a triangle can’t have two obtuse angle because the sum of all the angles of it can’t be more than ${180}^{\circ }$. It is always equal to ${180}^{\circ }$.

Solution

We know that sum of all the angles in a triangle is ${180}^{\circ }$.

The sum of all the angles is ${45}^{\circ }+{64}^{\circ }+{72}^{\circ }={181}^{\circ }$. So, we can’t draw any triangle having sum of all the angle ${181}^{\circ }$.

Solution

We know that sum of all the angles in a triangle is ${180}^{\circ }$.

Sum of these angles $={53}^{\circ }+{64}^{\circ }+{63}^{\circ }={180}^{\circ }$. So, we can draw infinitely many triangles having its angles as ${53}^{\circ },{64}^{\circ }$ and ${63}^{\circ }$.

Solution

See the given figure, $l\text{ and }m$ if a transversal intersects two parallel lines, then sum of interior angles on the same side of a transversal is supplementary.

$x+{44}^{\circ }={180}^{\circ }$

$\begin{array}{rl}& x={180}^{\circ }-{44}^{\circ }\\ & x={136}^{\circ }\end{array}$

Solution

No, because if it will be a right angle only when they form a linear pair.

Solution

If two intersecting line are formed right then by using linear pair axiom aniom, other three angles will be a right angle.

Solution

In the first figure, sum of two interior angle is: ${132}^{\circ }+{48}^{\circ }={180}^{\circ }$ [Equal to ${180}^{\circ }$ ]

Hence, we know that, if sum of two interior angle are equal on the same side of $n$ is ${180}^{\circ }$, then they are the parallel lines.

In the second figure, sum of two interior angle is:

${73}^{\circ }+{106}^{\circ }={179}^{\circ }\ne {180}^{\circ }$.

Hence, we know that, if sum of two interior angle are equal on the same side of $r$ is not equal to ${180}^{\circ }$, then they are not the parallel lines.

Solution

If two lines $l$ and $m$ are perpendicular to the same line $n$, then each of the two corresponding angles formed by these lines $l$ and $m$ with the line $n$ are equal to ${90}^{\circ }$.

Hence the line $l$ and $m$ are not perpendicular but parallel.

Solution

Given:

$OD$ is the bisector of $\mathrm{\angle }AOC,OE$ is the bisector of $\mathrm{\angle }BOC$ and $OD\perp OE$

To prove that point $A,O$ and $B$ are collinear that is $AOB$ are straight line.

$\mathrm{\angle }AOC=2\mathrm{\angle }DOC$

$\mathrm{\angle }COB=2\mathrm{\angle }COE$

Now, adding equations (I) and (II), get:

$\mathrm{\angle }AOC+\mathrm{\angle }COB=2\mathrm{\angle }DOC+\mathrm{\angle }COE$

$\mathrm{\angle }AOC+\mathrm{\angle }COB=2(\mathrm{\angle }DOC+\mathrm{\angle }COE)$

$\mathrm{\angle }AOC+\mathrm{\angle }COB=2\mathrm{\angle }DOC$

$\mathrm{\angle }AOC+\mathrm{\angle }COB=2\times {90}^{\circ }$

$\mathrm{\angle }AOC+\mathrm{\angle }COB={180}^{\circ }$

$\mathrm{\angle }AOC={180}^{\circ }$

So, $\mathrm{\angle }AOC+\mathrm{\angle }COB$ are forming linear pair or we can say that $AOB$ is a straight line.

Hence, point $A,O$ and $B$ are collinear.

Solution

See the given figure,

$\mathrm{\angle }5+\mathrm{\angle }6={180}^{\circ }$ (Linear pair angle)

$\mathrm{\angle }5+{120}^{\circ }={180}^{\circ }$

$\mathrm{\angle }5={180}^{\circ }-{120}^{\circ }$

$\mathrm{\angle }5={60}^{\circ }$

Then, $\mathrm{\angle }1=\mathrm{\angle }5[Each={60}^{\circ }]$

Since, these are corresponding angles.

Hence, the line $m$ and $n$ are parallel.

Solution

According to the question,

Line $l||m$ and $t$ is the transversal.

$\mathrm{\angle }MAB=\mathrm{\angle }SBA[$ Alt. $\mathrm{\angle }s]$

${\displaystyle \frac{1}{2}}\mathrm{\angle }MAB={\displaystyle \frac{1}{2}}\mathrm{\angle }SBA$

$\mathrm{\angle }PAB=\mathrm{\angle }QBA$

But, $\mathrm{\angle }PAB$ and $\mathrm{\angle }QBA$ are alternate angles.

Hence, $AP||BQ$.

Solution

See the given figure, $AP||BQ,AP$ and $BQ$ are the bisectors of alternate interior angles $\mathrm{\angle }CAB$ and $\mathrm{\angle }ABF$.

To show that $l||m$.

Now, prove that $AP||BQ$ are $t$ is transversal, therefore:

$\mathrm{\angle }PAB=\mathrm{\angle }ABQ$ [Alternate interior angle]

$2\mathrm{\angle }PAB=2\mathrm{\angle }ABQ$ [Multiplying both sides by 2 in equation (I)]

Since, alternate interior angle are equal.

So, if two alternate interior angle are equal then lines are parallel.

Hence, $l||m$.

Solution

According to the question:

Given:

Producing $DE$ to intersect $BC$ at $P$.

$EF||BC$ and $DP$ is the transversal,

$\mathrm{\angle }DEF=\mathrm{\angle }DPC$… (l) [Corresponding $\mathrm{\angle }s$ ]

See the above figure, $AB||DP$ and $BC$ is the transversal,

$\mathrm{\angle }DPC=\mathrm{\angle }ABC$… (II) [Corresponding $\mathrm{\angle }s$ ]

Now, from equation (I) and (II), get:

$\mathrm{\angle }ABC=\mathrm{\angle }DEF$

Hence, proved.

Solution

See in the figure, $BA||ED$ and $BC||EF$.

Show that $\mathrm{\angle }ABC+\mathrm{\angle }DEF={180}^{\circ }$.

Produce a ray PE opposite to ray EF.

Prove: $BC||EF$

Now, $\mathrm{\angle }EPB+\mathrm{\angle }PBC={180}^{\circ }$ [sum of co interior is ${180}^{\circ }$ ]

Now, $AB||ED$ and $PE$ is transversal line,

$\mathrm{\angle }EPB=\mathrm{\angle }DEF$ [Corresponding angles]

Now, from equation (I) and (II),

$\mathrm{\angle }DEF+\mathrm{\angle }PBC={180}^{\circ }$

$\mathrm{\angle }ABC+\mathrm{\angle }DEF={180}^{\circ }[$ Because $\mathrm{\angle }PBC=\mathrm{\angle }ABC]$

Hence, proved.

Solution

See in the given figure, $DE||QR$ and the line $n$ is the transversal line. $\mathrm{\angle }EAB+\mathrm{\angle }RBA={180}^{\circ }\dots$ (I) [The interior angles on the same side of transversal are supplementary.]

Now, $\mathrm{\angle }PAB+\mathrm{\angle }PBA={90}^{\circ }$

Then, from triangle APB, given: $\mathrm{\angle }APB={180}^{\circ }-(\mathrm{\angle }PAB+\mathrm{\angle }PBA)$

So, $\mathrm{\angle }APB={180}^{\circ }-{90}^{\circ }={90}^{\circ }$

Solution

Given in the question, ratio of angles is: $2:3:4$.

Let the angles of the triangle be $2x,3x$ and $4x$.

So,

$2x+3x+4x={180}^{\circ }$ [sum of angles of triangle is ${180}^{\circ }$ ]

$9x={180}^{\circ }$

$x={\displaystyle \frac{{180}^{\circ }}{9}}$

$x={20}^{\circ }$

Therefore, $2x=2\times {20}^{\circ }={40}^{\circ }$

$3x=2\times {20}^{\circ }={60}^{\circ }$

And, $4x=4\times {20}^{\circ }={80}^{\circ }$

Hence, the angle of the triangles are ${40}^{\circ },{60}^{\circ }$ and ${80}^{\circ }$.

Solution

Given:

In triangle $ABC$,

$\mathrm{\angle }A={90}^{\circ }$ and $AL\perp BC$

To prove: $\mathrm{\angle }BAL=\mathrm{\angle }ACB$

Proof: Let $\mathrm{\angle }ABC=x$

$\mathrm{\angle }BAL={90}^{\circ }-x$

As, $\mathrm{\angle }A=x$

$\mathrm{\angle }CAL=x$

$\mathrm{\angle }ABC=\mathrm{\angle }CAL$

$\mathrm{\angle }ABC=\mathrm{\angle }ACB$

Hence, proved.

Solution

According to the question:

Two line $p$ and $n$ are respectively perpendicular to two parallel line $l$ and $m$, that is $P\perp l$ and $n\perp m$.

To prove that $p$ is parallel to $n$.

Given: $n\perp m$

So, $\mathrm{\angle }1={90}^{\circ }$

Now, $P\perp l$

So, $\mathrm{\angle }2={90}^{\circ }$

Since, 1 is parallel to $m$. So,

$\mathrm{\angle }2=\mathrm{\angle }3$

[Corresponding $\mathrm{\angle }s$ ]

So,

$\mathrm{\angle }2={90}^{\circ }$

From equation (I) and (II), get:

$\mathrm{\angle }1=\mathrm{\angle }3$

[each ${90}^{\circ }$ ]

But these are corresponding angles.

Hence, $p||n$.

Solution

Two lines $AB$ and $CD$ intersect at point $O$.

To prove: (i) $\mathrm{\angle }AOC=\mathrm{\angle }BOD$

(ii) $\mathrm{\angle }AOD=\mathrm{\angle }BOC$

Proof: (i)

Ray on stands on line CD. So,

$\mathrm{\angle }AOC+\mathrm{\angle }AOD={180}^{\circ }\dots$ (I) [linear pair axiom]

Similarly, ray OD stands on line $AB$. So,

$\mathrm{\angle }AOD+\mathrm{\angle }BOD={180}^{\circ }$

Now, from equation (I) and (II), get:

$\mathrm{\angle }AOC+\mathrm{\angle }AOD=\mathrm{\angle }AOD+\mathrm{\angle }BOD$

$\mathrm{\angle }AOC=\mathrm{\angle }BOD$

Hence, proved.

(ii) Ray OD stands on line $AB$. $\mathrm{\angle }AOD+\mathrm{\angle }BOD={180}^{\circ }$ … (III) [Linear pair axiom]

Similarly, ray $OB$ stands on line $CD$. So,

$\mathrm{\angle }DOB+\mathrm{\angle }BOC={180}^{\circ }$

From equations (III) and (IV), get:

$\mathrm{\angle }AOD+\mathrm{\angle }BOD=\mathrm{\angle }DOB+\mathrm{\angle }BOC$

$\mathrm{\angle }AOD=\mathrm{\angle }BOC$

Hence, proved.

Solution

Given: in triangle $ABC$, produce $BC$ to $D$ and the bisectors of $\mathrm{\angle }ABC$ and $\mathrm{\angle }ACD$ meet at point T.

To prove that $\mathrm{\angle }BTC={\displaystyle \frac{1}{2}}\mathrm{\angle }BAC$

Proof: In triangle $ABC,\mathrm{\angle }ACD$ is an exterior angle.

$\mathrm{\angle }ACD=\mathrm{\angle }ABC+\mathrm{\angle }CAB$ [We know that exterior angle of a triangle is equal to the sum of two opposite angle]

${\displaystyle \frac{1}{2}}\mathrm{\angle }ACD={\displaystyle \frac{1}{2}}\mathrm{\angle }CAB+{\displaystyle \frac{1}{2}}\mathrm{\angle }ABC$ [Dividing both sides by 2 in the above equation]

$\mathrm{\angle }TCD={\displaystyle \frac{1}{2}}\mathrm{\angle }CAB+{\displaystyle \frac{1}{2}}\mathrm{\angle }ABC$

[Since, CT is the bisector of

$\mathrm{\angle }ACD$ that is $.{\displaystyle \frac{1}{2}}\mathrm{\angle }ACD=\mathrm{\angle }TCD]$

Now, in triangle BTC,

$\mathrm{\angle }TCD=\mathrm{\angle }BTC+\mathrm{\angle }CBT$ [We know that exterior angle of the triangle is equal to the sum of two opposite angles]

$\mathrm{\angle }TCD=\mathrm{\angle }BTC+{\displaystyle \frac{1}{2}}\mathrm{\angle }ABC$

…(II) [Since, BT is the bisector of triangle

$.ABC\mathrm{\angle }CBT={\displaystyle \frac{1}{2}}\mathrm{\angle }ABC]$

Now, from equation (I) and (II), get:

$\begin{array}{rl}{\displaystyle \frac{1}{2}}\mathrm{\angle }CAB+{\displaystyle \frac{1}{2}}\mathrm{\angle }ABC& =\mathrm{\angle }BTC+{\displaystyle \frac{1}{2}}\mathrm{\angle }ABC\\ {\displaystyle \frac{1}{2}}\mathrm{\angle }CAB& =\mathrm{\angle }BTC\\ {\displaystyle \frac{1}{2}}\mathrm{\angle }BAC& =\mathrm{\angle }BTC\end{array}$

Hence, proved.

Solution

Given: Lines DE $||QR$ and the line $DE$ intersected by transversal at $A$ and the line $QR$ intersected by transversal at $B$. Also, $BP$ and $AF$ are the bisector of angle $\mathrm{\angle }ABR$ and $\mathrm{\angle }CAE$ respectively.

To prove: $BP||FA$

Proof: DE $||QR$

$\mathrm{\angle }CAE=\mathrm{\angle }ABR$ [Corresponding angles]

${\displaystyle \frac{1}{2}}\mathrm{\angle }CAE={\displaystyle \frac{1}{2}}\mathrm{\angle }ABR$ [Dividing both side by 2 in the above equation]

$\mathrm{\angle }CAF=\mathrm{\angle }ABP$ [Since, bisector of angle $\mathrm{\angle }ABR$ and $\mathrm{\angle }CAE$ are BP and AF respectively]

Because these are the corresponding angles on transversal line $n$ and are equal.

Hence, BP $||$ FA.

Solution

Drawn a perpendicular line from the point $p$ as $PM\perp AB$. So, $\mathrm{\angle }PMB={90}^{\circ }$

Let if possible, drown another perpendicular line $PN\perp AB$. So, $\mathrm{\angle }PMB={90}^{\circ }$

Since, $\mathrm{\angle }PMB=\mathrm{\angle }PNB$ it will be possible when $PM$ and $PN$ coincide with each other.

Therefore, at a given point we can draw only one perpendicular to a given line.

Solution

Given:

Let lines $l$ and $m$ are two intersecting lines. Again, let $n\perp p$ to the intersecting lines meet at point $D$.

To prove that two lines $n$ and $p$ intersecting at a point.

Proof:

Let consider that line $n$ and $p$ are intersecting each other it means lines $n$ and $p$ are parallel to each other.

$n||p$

Therefore, lines $n$ and $p$ are perpendicular to $m$ and $l$ respectively.

Now, by using equation (I), $n|p$, it means that $l$ and $m$. it is a contradiction.

Since, our assumption is wrong.

Hence, line $n$ and $p$ are intersect at a point.

Solution

If triangle is an acute triangle then all the angle will be acute angle and sum of the all angle will be ${180}^{\circ }$.

If a triangle is a right angle triangle then one angle will be equal to ${90}^{\circ }$ and remaining two angle will be acute angles and sum of all the angles will be ${180}^{\circ }$.

Hence, a triangle must have at least two acute angles.

Solution

Given in triangle $PQR,\mathrm{\angle }Q>\mathrm{\angle }R$, $PA$ is the bisector of $\mathrm{\angle }QPR$ and $PM\perp QR$.

To prove that $\mathrm{\angle }APM={\displaystyle \frac{1}{2}}(\mathrm{\angle }Q-\mathrm{\angle }R)$

Proof: PA is the bisector of $\mathrm{\angle }QPR$. So,

$\mathrm{\angle }QPA=\mathrm{\angle }APR$

In angle $PQM,\mathrm{\angle }Q+\mathrm{\angle }PMQ+\mathrm{\angle }QPM={180}^{\circ }$

(I) [Angle sum property of a triangle]

$\mathrm{\angle }Q+{90}^{\circ }+\mathrm{\angle }QPM={180}^{\circ }[\mathrm{\angle }PMR={90}^{\circ }]$

$\mathrm{\angle }Q={90}^{\circ }-\mathrm{\angle }QPM$

In triangle PMR, $\mathrm{\angle }PMR+\mathrm{\angle }R+\mathrm{\angle }RPM={180}^{\circ }$ [Angle sum property of a triangle]

${90}^{\circ }+\mathrm{\angle }R+\mathrm{\angle }RPM={180}^{\circ }[\mathrm{\angle }PMR={90}^{\circ }]$

$\mathrm{\angle }R={180}^{\circ }-{90}^{\circ }-\mathrm{\angle }RPM$

$\mathrm{\angle }R={180}^{\circ }-{90}^{\circ }-\mathrm{\angle }RPM$

$\mathrm{\angle }R={90}^{\circ }-\mathrm{\angle }RPM$

Subtracting equation (III) from equation (II), get:

$$\begin{array}{rl}& \mathrm{\angle }Q-\mathrm{\angle }R=({90}^{\circ }-\mathrm{\angle }APM)-({90}^{\circ }-\mathrm{\angle }RPM)\\ & \mathrm{\angle }Q-\mathrm{\angle }R=\mathrm{\angle }RPM-\mathrm{\angle }QPM\\ & \mathrm{\angle }Q-\mathrm{\angle }R=(\mathrm{\angle }RPA+\mathrm{\angle }APM)-(\mathrm{\angle }QPA-\mathrm{\angle }APM)\dots (IV)\\ & \mathrm{\angle }Q-\mathrm{\angle }R=\mathrm{\angle }QPA+\mathrm{\angle }APM-\mathrm{\angle }QPA+\mathrm{\angle }APM\text{ [As, }\mathrm{\angle }RPA=\mathrm{\angle }QPA]\\ & \mathrm{\angle }Q-\mathrm{\angle }R=2\mathrm{\angle }APM\\ & \mathrm{\angle }APM={\displaystyle \frac{1}{2}}(\mathrm{\angle }Q-\mathrm{\angle }R)\end{array}$$

Hence, proved.