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MATHS 5 Marks Questions

Quadrilaterals · Rajasthan Board (RBSE) STD 9 (Rajasthan - English Medium). 15 questions.

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Question 15 Marks
The line segment joining the mid-points of two sides of a triangle is parallel to the third side.
Answer

Image
You can prove this theorem using the following clue:
Observe Fig 8.16 in which $\mathrm{E}$ and $\mathrm{F}$ are mid-points of $\mathrm{AB}$ and $\mathrm{AC}$ respectively and $\mathrm{CD} \| \mathrm{BA}$.
$\triangle \mathrm{AEF} \cong \triangle \mathrm{CDF} \quad \text { (ASA Rule) }$
So, $\mathrm{EF}=\mathrm{DF}$ and $\mathrm{BE}=\mathrm{AE}=\mathrm{DC} \quad$ (Why?)
Therefore, BCDE is a parallelogram. (Why?)
This gives $\mathrm{EF} \| \mathrm{BC}$.
In this case, also note that $\mathrm{EF}=\frac{1}{2} \mathrm{ED}=\frac{1}{2} \mathrm{BC}$.
Can you state the converse of Theorem 8.8? Is the converse true?
You will see that converse of the above theorem is also true which is stated as below:
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Question 25 Marks
If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.
Answer
Image
Can you reason out why?
Let sides $A B$ and $C D$ of the quadrilateral $A B C D$ be equal and also $A D=B C$ (see Fig. 8.3). Draw diagonal AC.
Clearly, $\triangle \mathrm{ABC} \cong \triangle \mathrm{CDA}$ (Why?)
So, $\angle \mathrm{BAC}=\angle \mathrm{DCA}$
and $\angle \mathrm{BCA}=\angle \mathrm{DAC}$ (Why?)
Can you now say that $A B C D$ is a parallelogram? Why?
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Question 35 Marks
In a parallelogram, opposite sides are equal.
Answer
You have already proved that a diagonal divides the parallelogram into two congruent triangles; so what can you say about the corresponding sides? They are equal. gparts say, 1 the corresponding
So, AB DC and AD = BC
Now what is the converse of this result? You already know that whatever is given in a theorem, the same is to be proved in the converse and whatever is proved in the theorem it is given in the converse. Thus, Theorem 8.2 can be stated as given below:
If a quadrilateral is a parallelogram, then each pair of its opposite sides is equal. So its converse is:
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Question 45 Marks
A diagonal of a parallelogram divides it into two congruent triangles.
Answer
Proof : Let $\mathrm{ABCD}$ be a parallelogram and $\mathrm{AC}$ be a diagonal (see Fig. 8.2). Observe that the diagonal $\mathrm{AC}$ divides parallelogram $\mathrm{ABCD}$ into two triangles, namely, $\triangle \mathrm{ABC}$ and $\triangle \mathrm{CDA}$. We need to prove that these triangles are congruent.
Image
In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{CDA}$, note that $\mathrm{BC} \| \mathrm{AD}$ and $\mathrm{AC}$ is a transversal.
So, $\quad \angle \mathrm{BCA}=\angle \mathrm{DAC}$ (Pair of alternate angles)
Also, $\mathrm{AB} \| \mathrm{DC}$ and $\mathrm{AC}$ is a transversal.
So, $\quad \angle \mathrm{BAC}=\angle \mathrm{DCA}$ (Pair of alternate angles)
and $\mathrm{AC}=\mathrm{CA} \quad$ (Common)
So, $\triangle \mathrm{ABC} \cong \triangle \mathrm{CDA} \quad$ (ASA rule)
or, diagonal $\mathrm{AC}$ divides parallelogram $\mathrm{ABCD}$ into two congruent triangles $\mathrm{ABC}$ and $\mathrm{CDA}$.
Now, measure the opposite sides of parallelogram $A B C D$. What do you observe?
You will find that $\mathrm{AB}=\mathrm{DC}$ and $\mathrm{AD}=\mathrm{BC}$.
This is another property of a parallelogram stated below:
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Question 55 Marks
$\text{ABC}$ is a triangle right angled at $C.$ A line through the mid$-$point $M$ of hypotenuse $AB $ and parallels to $BC$ intersects $AC$ at $D$ Show that :
  1. $D$ is the mid$-$point of $AC$
  2. $MD \perp AC$
  3. $CM = MA =  \frac{1}{2} AB$
Answer
Given: $ABC$ is a triangle right angled at $C.$ A line through the mid$-$point $M$ of hypotenuse $AB$ and parallels to $BC$ intersects $AC$ at $D.$

To Prove :
  1. $D$ is the mid$-$point of $AC (ii) MD \perp AC$
  2. $CM = MA = \frac{1}{2}AB$
Proof :
  1. In $ACB,$
    As $M$ is the mid$-$point of $AB$ and $MD \ || \ BC$
    $\therefore D$ is the mid$-$point of $AC . . . [$By converse of mid$-$point theorem$]$
  2. As $MD \ || \ BC$ and $AC$ intersects them
    $\angle ADM = \angle ACB . . . [$Corresponding angles$]$
    But $\angle ACB = 90^\circ . . .[$Given$]$
  3. $\therefore \angle ADM = 90^\circ \Rightarrow MD \perp AC$
  4. Now $\angle ADM + \angle CDM = 180^\circ . . .[$Linear pair axiom$]$
    $\angle ADM = \angle CDM = 90^\circ$
    In $\triangle ADM$ and $\triangle CDM$
    $AD = CD . . . [$As $D$ is the mid$-$point of $AC]$
    $\angle ADM = \angle CDM . . .[$Each $90^\circ ]$
    $DM = DM . . .[$Common$]$
    $\therefore \triangle ADM \cong \triangle CDM . . .[$By $SAS$ rule$]$
    $\therefore MA = MC . . .[c.p.c.t.]$
    But $M$ is the mid$-$point of $AB$
  5. $\therefore MA = MB = \frac{1}{2}AB$
    $\therefore MA = MC = \frac{1}{2}AB$
    $\therefore CM = MA = \frac{1}{2}AB$
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Question 65 Marks
$\text{ABCD}$ is a rectangle and $\text{P, Q, R}$ and $\text{S}$ are mid$-$points of the sides $\text{AB, BC, CD}$ and $\text{DA}$ respectively. Show that quadrilateral $\text{PQRS}$ is a rhombus.
Answer
$\text{ABCD}$ is a rectangle. $P, Q, R$ and S are the mid$-$points of $AB, BC, CD$ and $DA$ respectively. $PQ, QR, RS$ and $SP$ are joined.
To Prove : Quadrilateral $\text{PQRS}$ is a rhombus.
Construction: Join $AC.$

Proof: In $\triangle ABC,$
As $P$ and $Q$ are the mid$-$points of $AB$ and $BC$ respectively.
$PQ \ || \ AC$ and $PQ = \frac{1}{2} AC ...(1)$
In $\triangle ADC,$
As $S$ and $R$ are the midpoints of $AD$ and $DC$ respectively.
$SR \ || \ AC$ and $SR = \frac{1}{2} AC ...(2)$
From $(1)$ and $(2)$
$PQ \ || \ SR$ and $PQ = SR$
$\therefore$ Quadrilateral $\text{PQRS}$ is a parallelogram $. . . (3)$
In rectangle $\text{ABCD,}$
$AD = BC . . .[$Opposite sides$]$
$\therefore \frac{1}{2} AD = \frac{1}{2} BC ...[$Halves of equals are equal$]$
$\therefore AS = BQ$
In $\triangle APS$ and $\triangle BPQ$
As $P$ is the mid$-$point of $AB$
$AP = BP$
$AS = BQ . . . [$As proved above$]$
$\angle PAS = \angle PBQ . . .[$Each $90^\circ ]$
$\therefore \triangle APS \cong \triangle BPQ . . . [$By $SAS$ axiom$]$
$\therefore PS = PQ . . .[c.p.c.t.] . . . (4)$
According to $(3), (4) PQRS$ is a rhombus.
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Question 75 Marks
$\text{ABCD}$ is a rhombus and $\text{P, Q, R}$ and $\text{S}$ are the mid$-$points of the sides $\text{AB, BC, CD}$ and $\text{DA}$ respectively. Show that the quadrilateral $\text{PQRS}$ is a rectangle.
Answer
Given: $ \text{ABCD}$ is a rhombus, $ \text{P, Q, R, S}$ are the mid$-$points of $ \text{AB, BC, CD, DA}$ respectively. $ \text{PQ, QR, RS}$ and $ \text{SP}$ are joined.
To Prove: $ \text{PQRS}$ is a rectangle.
Construction: Join $AC$ and $BD.$

Proof : In $\triangle RDS$ and $\triangle PBQ$
$DS = QB . . .[$Halves of opp. sides of $|| \ gm$ are equal$]$
$DR = PB . . .[$Halves of opp. sides of $|| \ gm$ are equal$]$
$\angle SDR = \angle QBP . . .[$Opp. $\angle$ of $|| \ gm$ are equal$]$
$\therefore \triangle RDS \cong\triangle PBQ . . . . [c.p.c.t.]$
$\therefore SR = PQ$
In $\triangle RCQ$ and $\triangle PAS$
$RC = AP . . .[$Halves of opp. sides of $|| \ gm$ are equal$]$
$CQ = AS . . .[$Halves of opp. sides of $|| \ gm$ are equal$]$
$\angle RCQ = \angle PAS . . .[$Opp. $\angle$ of $|| \ gm$ are equal$]$
$\therefore \triangle RCQ \cong \triangle PAS . . . . [c.p.c.t.]$
$\therefore RQ = SP$
$\therefore$ In $ \text{PQRS},$
$SR = PQ$ and $RQ = SP$
$\therefore \text{PQRS}$ is a parallelogram,
In $\triangle CDB,$
As $R$ and $Q$ are the mid$-$points of $DC$ and $CB$ respectively.
$\triangle RQ \ || \ DB \therefore  RF \ || \ EO $
Similarly, $RE \ || \ FO$
$\therefore \text{OFRE}$ is $a \ ||  \ gm$
As opp. $\angle$ of $|| \ gm$ are equal and diagonals of rhombus intersect at $90^\circ$
$\therefore R = EOF = 90^\circ$
Thus $\text{PQRS}$ is a rectangle.
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Question 85 Marks
ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA. AC is a diagonal. Show that
  1. SR || AC and SR = $\frac{1}{2}$AC
  2. PQ = SR
  3. PQRS is a parallelogram.
Answer
Given: ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA. AC is a diagonal.
To Prove :

  1. SR || AC and SR = $\frac{1}{2}$AC
  2. PQ = SR
  3. PQRS is a parallelogram

Proof :

  1. In  $\triangle$DAC,
    As S is the mid-point of DA and R is the mid-point of DC
    $\therefore$ SR || AC and SR = $\frac{1}{2}$AC . . . [Mid point theorem]
  2. In $\triangle$BAC,
    As P is the mid-point of AB and Q is the mid-point of BC
    $\therefore$ PQ || AC and PQ = $\frac{1}{2}$AC . . . [Mid point theorem]
    But from (i) SR = $\frac{1}{2}$AC
    $\therefore$ PQ = SR
  3. PQ || AC . . .[From (i)]
    SR || AC . . .[From (i)]
    $\therefore$ PQ || SR . . .[Two lines parallel to the same line are parallel to each other]
    Similarly, PQ = SR . . .[From (ii)]
    $\therefore$ PQRS is a parallelogram . . . [A quadrilateral is a parallelogram if a pair of opposite sides is parallel and is of equal length]
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Question 95 Marks
In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ.

Show that
  1. $\triangle$APD $\cong$$\triangle$CQB
  2. AP = CQ
  3. $\triangle$AQB $\cong$$\triangle$CPD
  4. AQ = CP
  5. APCQ is a parallelogram.
Answer
In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ.
To Prove :

  1. $\triangle$APD $\cong$ $\triangle$CQB
  2. AP = CQ
  3. $\triangle$AQB $\cong$ $\triangle$CPD
  4. AQ = CP
  5. APCQ is a parallelogram.

Construction: Join AC to intersect BD at O.

  1. In $\triangle$APD and $\triangle$CQB
    AD || BC . . . [Opp. sides of || gm ABCD]
    A transversal BD intersects them
    $\therefore$ $\angle$ADB = $\angle$CBD . . .[Alternate angles]
    $\Rightarrow$ $\angle$ADP = $\angle$CBQ . . . (1)
    DP = BQ . . . [Given] . . . (2)
    AD = CB . . . [Opp. sides of || gm ABCD] . . . (3)
    According to (1), (2), (3)
    $\triangle$APD $\cong$$\triangle$CQB
  2. $\triangle$APD $\cong$$\triangle$CQB . . . [As proved above]
    $\therefore$ AP = CQ . . . [c.p.c.t.]
  3. In $\triangle$AQB and $\triangle$CPD,
    AB || CD . . . . . . [Opp. sides of || gm ABCD]
    A transversal BD intersects them
    $\therefore$ $\angle$ABD = $\angle$CDP . . .[Alternate angles]
    $\Rightarrow$  $\angle$ABQ = $\angle$CDP
    AQ = CP . . . .[From (3)]
    QB = PD . . . .[Given]
    AB = CD . . . .[Opp. sides of || gm ABCD]
    $\therefore$ $\triangle$AQB $\cong$$\triangle$CPD . . . [By SAS rule]
  4. $\triangle$AQB $\cong$ $\triangle$CPD . . . [As proved above]
    $\therefore$ AQ = CP . . . [c.p.c.t]
    As the diagonals of a parallelogram bisect each other
    $\therefore$ OB = OD
    $\therefore$ OB – BQ = OD – DP . . .[Given : BQ = DP]
    $\therefore$ OQ = OP . . . (1)
    Also OA = OC . . . [As diagonals of a || gm bisect each other] . . .(2)
    In view of (1) and (2) APCQ is a parallelogram.
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Question 105 Marks
ABCD is a rectangle in which diagonal AC bisects $\angle$A as well as $\angle$C. Show that:
  1. ABCD is a square
  2. diagonal BD bisects $\angle$B as well as $\angle$D.
Answer


Given : ABCD is a rectangle,
$\angle$A = $\angle$C
$\frac{1}{2} \angle A=\frac{1}{2} \angle C$ 
To prove: ABCD is a square proof:
  1. $\angle$DAC = $\angle$DCA (AC bisects A and C)
    CD = DA (Sides opposite to equal angles are also equal)
    However,
    DA = BC and AB = CD (Opposite sides of a rectangle are equal)
    AB = BC = CD = DA
    ABCD is a rectangle and all of its sides are equal.
    Hence, ABCD is a square
  2. Let us join BD
    In $\triangle$BCD,
    BC = CD (Sides of a square are equal to each other)
    $\angle$CDB = $\angle$CBD (Angles opposite to equal sides are equal)
    However,
    $\angle$CDB = $\angle$ABD (Alternating interior angles for AB $\parallel$ CD)
    $\angle$CBD = $\angle$ABD
    BD bisects $\angle$B
    Also,
    $\angle$CDB = $\angle$ABD
    BD bisects $\angle$D.
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Question 115 Marks
Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
Answer
Given: The diagonals $AC$ and $BD$ of a quadrilateral $\text{ABCD}$ are equal and bisect each other at right angles.

To Prove : Quadrilateral $\text{ABCD}$ is a square.
Proof : In $\triangle OAD$ and $\triangle OCB,$
$OA = OC, OD = OB . . . [$Given, As diagonals $AC$ and $BD$ of a quadrilateral $\text{ABCD}$ are equal and bisect each other $]$
$\angle AOD = \angle COB . . . [$Vertically opposite angles$]$
$\therefore \triangle OAD \cong \triangle OCB . . . [$By $\text{SAS}$ rule$]$
$\triangle AD = CB . . . [c.p.c.t.]$
$\angle ODA = \angle OBC . . . [c.p.c.t.]$
$\therefore AD || BC$
As $AD = CB$ and $AD || CB$
$\therefore$ Quadrilateral $\text{ABCD}$ is a $|| gm.$
In $\triangle AOB$ and $\triangle AOD,$
$AO = AO . . . [$Common$]$
$OB = OD . . . [$Given$]$
$\angle AOB = \angle AOD . . . [$Given : Each $90^\circ]$
$\therefore \triangle AOB \cong \triangle AOD . . . [$By $\text{SAS}$ rule$]$
$\therefore AB = AD$
As $\text{ABCD}$ is a parallelogram and $AB = AD$
$\therefore \text{ABCD}$ is a rhombus.
Again in $\triangle ABC$ and $\triangle BAD$
$AC = BD$
$BC = AD . . . [$Given$]$
$AB = BA . . . [$Common$]$
$\therefore \triangle ABC \cong \triangle BAD . . . [$By $\text{SSS}$ rule$]$
$\therefore \angle ABC = \angle BAD . . . [c.p.c.t.]$
As $AD || BC . . . [$Opp. sides of $|| gm\  \text{ABCD}]$
and transversal $AB$ intersects them.
$\angle ABC + \angle BAD = 180^\circ . . .[$Sum of interior angles on the same side of transversal$]$
$\therefore \angle ABC = \angle BAD = 90^\circ$
Also, $\angle BCD = \angle ADC = 90^\circ$
$\therefore \text{ABCD}$ is a square.
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Question 125 Marks
Show the diagonals of a square are equal and bisect each other at right angles.
Answer
Given: $A$ square $\text{ABCD}.$

To Prove : $(i) AC = BD$ and
$(ii)$ Diagonals bisect each other at right angles.
Proof :
  1. In $\triangle ADB$ and $\triangle BCA, $we have
    $AD = BC ...[$As sides of a square are equal$]$
    $\angle BAD = \angle ABC ...[$All interior angles are of 90^\circ$]$
    $AB = BA ...[$Common$]$
    $\triangle ADB \cong$
    $\triangle BCA ...[$By $\text{SAS}$ rule$]$
  2. $AC = BD ...[c.p.c.t.]$
  3. Now in $\triangle AOB$ and $\triangle COD, $we have
    $AB = CD ...[$Sides of a square$]$
    $\angle AOB = \angle COD ...[$Vertically opp. angles$]$
    $\angle OBA = \angle ODC ...[$Alternate interior angles are equal$]$
    $\triangle AOB \cong$
    $\triangle COD ...[$By $\text{ASA}$ rule$]$
    $OA = OC$ and $OB = OD ...[c.p.c.t.] ...(1)$
    Now consider $\triangle AOD$ and $COD.$
    $AD = CD ...[$Sides of square$]$
    $OA = OC ...[$As proved above$]$
    $OD = OD ...[$Common$]$
    $\triangle AOD \cong$
    $\triangle COD ...[$By $\text{SSS}$ rule$]$
    $\angle AOD = \angle COD ...[c.p.c.t.]$
    But $\angle AOD + \angle COD = 180^\circ ...[$linear pair$]$
    or $\angle AOD + \angle AOD = 180^\circ ...[$As $\angle AOD = \angle COD]$
    or $2\angle AOD = 180^\circ \therefore \angle AOD = 90^\circ ...(2)$
    From equation $(1)$ and $(2)$ it is clear that diagonals of a square bisect each other at right angles.
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MCQ 135 Marks
$\text{ABCD}$ is a trapezium in which $AB || CD$ and $AD = BC$.
Show that :
  • A
    $\angle A = \angle B$
  • B
    $\angle C = \angle D$
  • C
    $\triangle ABC \cong \angle BAD$
  • ✓
    diagonal $AC =$ diagonal $BD.$
Answer
Correct option: D.
diagonal $AC =$ diagonal $BD.$
Given : $\text{ABCD}$ is a trapezium in which $AB || CD$ and $AD = BC.$
To Prove :
  1. $\angle A = \angle B$
  2. $\angle C = \angle D$
  3. $\triangle ABC \cong \triangle BAD$
  4. diagonal $AC =$ diagonal $BD$.
Construction: Extend $AB$ and draw a line through $C$ parallel to $DA$ intersecting $AB$ produced at $E.$

Proof :
  1. $AB || CD . . . [$Given$]$
    $AD || ED . . .[$By construction$]$
    $\therefore AECD$ is a parallelogram $. . . [$A quadrilateral is a parallelogram if a pair of Opp. sides are parallel and of equal length$]$
    $\therefore AD = EC . . .[$Opp. sides of a $||\  gm$ are equal$]$
    But  $AD = BC . . .[$Given$]$
    $\therefore EC = BC$
    $\therefore \angle CBE = \angle CEB . . . [ \angle$s opposite to equal side of a triangle are equal$] . . .(1)$
    $\angle B + \angle CBE = 180^\circ . . .[$Linear pair axiom$]. . . (2)$
    As $AD || EC . . .[$By construction$]$
    and transversal $AE$ intersect them
    $\angle A + \angle CEB = 180^\circ . . . [$The sum of interior angles on the same side of the transversal is $180^\circ] . . . (3)$
    From $(2)$ and $(3),$
    $\angle B + \angle CBE = \angle A + \angle CEB$
    But $\angle CBE = \angle CEB . . .[$ From $(1)]$
    $\therefore \angle B = \angle A$ or $\angle A = \angle B$
  2. As $AB || CD$
    $\therefore \angle A + \angle D = 180^\circ. . .[$The sum of interior angles on the same side of the transversal is $180^\circ]$
    and $\angle B + \angle C = 180^\circ$
    $\therefore \angle A + \angle D = \angle B + \angle C$
    But $\angle A =  \angle B . . .[$As proved in $(i)]$
    $\therefore \angle D = \angle C$ or $\angle C =  \angle D$
  3. In $\triangle ABC$ and $\triangle BAD,$
    $AB = BA . . . [$Common$]$
    $BC = AD . . .[$Given$]$
    $\angle ABC = \angle BAD . . .[$ From $(i)]$
    $\therefore \triangle ABC \cong \triangle BAD . . .[\text{SAS}$ rule$]$
  4. As $\triangle ABC \cong \triangle BAD . . [$From $(iii)]$
    $\therefore AC = BD . . .[c.p.c.t.]$
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Question 145 Marks
In $\triangle$ABC and $\triangle$DEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively.

Show that:
  1. Quadrilateral ABED is a parallelogram
  2. Quadrilateral BECF is a parallelogram
  3. AD || CF and AD = CF
  4. quadrilateral ACFD is a parallelogram
  5. AC = DF
  6. $\triangle$ABC $\cong$$\triangle$DEF.
Answer
Given : In $\triangle$ABC and $\triangle$DEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively.
To prove:

  1. Quadrilateral ABED is a parallelogram
  2. Quadrilateral BECF is a parallelogram
  3. AD || CF and AD = CF
  4. quadrilateral ACFD is a parallelogram
  5. AC = DF
  6. $\triangle$ABC $\cong$$\triangle$DEF.

Proof :

  1. In quadrilateral ABED
    AB = DE and AB || DE . . .[Given]
    $\therefore$ quadrilateral ABED is a parallelogram . . .[As a quadrilateral is a parallelogram if a pair of opposite sides is of equal length]
  2. In quadrilateral BECF
    BC = EF and BC || EF . . .[Given]
    $\therefore$ quadrilateral BECF is a parallelogram . . .[As a quadrilateral is a parallelogram if a pair of opposite sides is of equal length]
  3. As ABED is a parallelogram . . .[As proved in (i)]
    $\therefore$ AD || BE and AD = BE . . .[As opp. sides of || gm are parallel and equal]. . . .(1)
    As AEFC is a parallelogram . . .[As proved in (ii)]
    $\therefore$ BE || CF and BE = CF . . .[As opp. sides of || gm are parallel and equal]. . . .(2)
    AD || CF and AD = CF . . . [From (1) and (2)]
  4. In quadrilateral ACFD,
    AD || CF and AD = CF . . .[From (iii)]
    $\therefore$ quadrilateral ACFD is a parallelogram . . .[As a quadrilateral is a parallelogram if a pair of opposite sides is of equal length]
  5. ACFD is a parallelogram
    $\therefore$ AC || DF and AC = DF . .[In a parallelogram opposite sides are parallel and of equal length]
  6. In $\Delta$ABC and $\Delta$DEF,
    AB = DE . . . [As ABED is a parallelogram]
    BC = EF . . . [As BEFC is a parallelogram]
    AC = DF . . .[As proved in (v)]
    $\therefore$ $\triangle$ABC$\cong$ $\triangle$DEF . . .[By SSS rule]
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Question 155 Marks
Two parallel lines $l$ and $m$ are intersected by a transversal $p$. Show that the quadrilateral formed by the bisectors of interior angles is a rectangle.
Answer

Here, it is given that $QR \parallel PS$ and transversal $p$ intersects them at points $A$ and $C$ respectively.
The bisectors of $\angle ACR$ and $\angle SAC$ intersect at $D$ and the bisectors of $\angle PAC$ and $\angle ACQ$ intersect at $B.$
We have to prove that quadrilateral$\text{ABCD}$ is a rectangle.
From the given figure, we have
$\angle PAC = \angle ACR [$Alternate angles as $l \parallel m$ and $p$ is a transversal$]$
So, $\frac{1}{2} \angle PAC = \frac{1}{2} \angle ACR$
i.e., $\angle BAC = \angle ACD$
These form a pair of alternate angles for lines $AB$ and $DC$ with $AC$ as a transversal and they are equal also.
So$, AB \parallel DC$
Similarly $, BC \parallel AD [$Considering $\angle ACB$ and $\angle CAD]$
Therefore, quadrilateral $\text{ABCD}$ is a parallelogram
Also, $\angle PAC + \angle CAS = 180^\circ [$ by the property of Linear pair$]$
So, $\frac{1}{2} \angle PAC + \frac{1}{2} \angle CAS = \frac{1}{2} \times 180^\circ = 90^\circ$
or, $\angle BAC + \angle CAD = 90^\circ$
or, $\angle BAD = 90^\circ$
So, $\text{ABCD}$ is parallelogram in which one angle $(\angle BAD )$ is $90^\circ$.
Therefore, $\text{ABCD}$ is a rectangle.
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