Please solve these two questions.
Of class 9 maths chapter 8- Quadrilaterals
Answers
9. Given, ABCD is a parallelogram and P and Q are points on BD such that DP = BQ
(i) We have to show,
∆APD ≅ ∆CBQ
In ∆APD and ∆CBQ, we have
DP = BQ [given]
AD = BC
[opposite sides are equal in parallelogram]
∠ADP = ∠CBQ
[since, AD || BC and BD is a transversal, so alternate interior angles are equal]
(ii) Since, ∆APD ≅ ∆CBQ
(iii) We have to show, ∆AQB ≅ ∆CPD
Now, in ∆AQB and ∆CPD, we have
BQ = DP [given]
AB = CD [opposite sides of parallelogram]
∠ABQ = ∠CDP
[since, AB || CD and BD is a transverse, so alternate interior angles]
(iv) Since, ∆ABQ ≅ ∆CPD
(v) We have to show, APCQ is a parallelogram.
Now, in ∆APQ and ∆CQP, we have
AQ = CP [from part (iv)]
AP = CQ [from part (ii)]
PQ = QP [common side]
∴ ∆APQ ≅ ∆CQP [by SSS congruence rule]
Then, ∠APQ = ∠CQP [by CPCT]
and ∠AQP = ∠CPQ [by CPCT]
Now, these equal angles form a pair of alternate angles, when line segment AP and QC are intersected by a transversal PQ.
∴ AP || QC and AQ || PC
Now, both pairs of opposite sides of quadrilateral APCQ are parallel and equal.
10. Given, ABCD is a parallelogram and AP and CQ are perpendicular from vertices A and C on diagonal BD.
Since, AB || DC and BD is a transversal.
∴ ∠CQD = ∠ABD ...(i)
(i) Now, in ∆CQD and ∆APB, we have
CD = BA [sides of a parallelogram]
∠CQD = ∠ABD = 90°
[since, AP ⊥ BD and CQ ⊥ BD, given]
∠CDQ = ∠ABP [from Eq. (i)]
(ii) Since, ∆APB ≅ ∆CQD
Answer:
Given, ABCD is a parallelogram and P and Q are points on BD such that DP = BQ
(i) We have to show,
∆APD ≅ ∆CBQ
In ∆APD and ∆CBQ, we have
DP = BQ [given]
AD = BC
[opposite sides are equal in parallelogram]
∠ADP = ∠CBQ
[since, AD || BC and BD is a transversal, so alternate interior angles are equal]
\small\sf \bf \pink{ ∴ ∆APD ≅ ∆CBQ\: [By \: SAS \: congruence \: rule]\: \: \:}∴∆APD≅∆CBQ[BySAScongruencerule]
(ii) Since, ∆APD ≅ ∆CBQ
\small\sf \bf \blue{ = > AP = CQ \: [By \: CPCT]\: \:} \:=>AP=CQ[ByCPCT]
(iii) We have to show, ∆AQB ≅ ∆CPD
Now, in ∆AQB and ∆CPD, we have
BQ = DP [given]
AB = CD [opposite sides of parallelogram]
∠ABQ = ∠CDP
[since, AB || CD and BD is a transverse, so alternate interior angles]
\small\sf \bf \red{ So, ∆ABQ \:≅ \: ∆CPD \: [By \: SAS \: congruence \: rule]\: \:}So,∆ABQ≅∆CPD[BySAScongruencerule]
(iv) Since, ∆ABQ ≅ ∆CPD
\small\sf \bf \purple{ ∴ AQ = CP\:[By CPCT] \:}∴AQ=CP[ByCPCT]
(v) We have to show, APCQ is a parallelogram.
Now, in ∆APQ and ∆CQP, we have
AQ = CP [from part (iv)]
AP = CQ [from part (ii)]
PQ = QP [common side]
∴ ∆APQ ≅ ∆CQP [by SSS congruence rule]
Then, ∠APQ = ∠CQP [by CPCT]
and ∠AQP = ∠CPQ [by CPCT]
Now, these equal angles form a pair of alternate angles, when line segment AP and QC are intersected by a transversal PQ.
∴ AP || QC and AQ || PC
Now, both pairs of opposite sides of quadrilateral APCQ are parallel and equal.
\small\sf \bf \orange{ Hence, \: APCQ \: is \: a \: parallelogram. \: Hence \: proved.\:} < /p > < p >Hence,APCQisaparallelogram.Henceproved.</p><p>
10. Given, ABCD is a parallelogram and AP and CQ are perpendicular from vertices A and C on diagonal BD.
Since, AB || DC and BD is a transversal.
∴ ∠CQD = ∠ABD ...(i)
(i) Now, in ∆CQD and ∆APB, we have
CD = BA [sides of a parallelogram]
∠CQD = ∠ABD = 90°
[since, AP ⊥ BD and CQ ⊥ BD, given]
∠CDQ = ∠ABP [from Eq. (i)]
\small\sf \bf \green{ ∴ ∆CQD ≅ ∆APB \: [by \: AAS \: congruence \: rule]\:} < /p > < p >∴∆CQD≅∆APB[byAAScongruencerule]</p><p>
(ii) Since, ∆APB ≅ ∆CQD
\small\sf \bf \pink{ ∴ AP = CQ \: [by CPCT]\:}∴AP=CQ[byCPCT]