Question

in a parallelogram ABCD two points P and Q are taken on diagonal BD such that DP=BQ show that
1.) triangle APD congruent to triangle CQB
2.) AP=CQ
3.) triangle AQB congruent to triangle CPD
4.)AQ=CP
5.)APCQ is a parallelogram
for class 9 NCERT exercise 8.1 chapter quadrilateral

Answer 1

Hello Mate!

1. In ∆APD and ∆BQC

AD = BC [ Since ABCD is a ||gm ]

PD = BQ [ Given ]

< ADP = < BQC [ Alternate angles ]

Hence ∆APD ~ ∆BQC by SAS congruency.

2. AP = QC [ c.p.c.t ]

3. In ∆AQB and ∆CDP

CD = AB [ Since ABCD is a ||gm ]

PD = BQ [ Given ]

< ABQ = < CDP [ Alternate angles ]

Hence ∆AQB ~ ∆CDP by SAS congruency.

4. AQ = CP [ c.p.c.t ]

5. In ∆APQ and ∆CQP we have,

AQ = CP

AP = QC

PQ = PQ

Hence ∆APQ ~ ∆CQP by SSS congruency.

< CPQ = < AQP and < CQP = < APQ [ c.p.c.t ]

Hence, AQ || CP and AP || QC.

Hence AQCP is a ||gm.

Have great future ahead!

Answer 2

here is your answer OK ☺☺☺☺☺

Given : ABCD is a parallelogram and P and Q are points on diagonal BD such that DP = BQ.


To Prove :

(i) ∆APD ≅ ∆CQB
(ii) AP = CQ
(iii) ∆AQB ≅ ∆CPD
(iv) AQ = CP
(v) APCQ is a parallelogram


Proof :

(i) In ∆APD and ∆CQB,we have
AD = BC [Opposite sides of a ||gm]
DP = BQ [Given]
∠ADP = ∠CBQ [Alternate angles]
∴ ∆APD ≅ ∆CQB [By SAS congruence]


(ii) ∴ AP = CQ [By CPCT]


(iii) In ∆AQB and ∆CPD, we have


AB = CD [Opposite sides of a ||gm]
DP = BQ [Given]
∠ABQ = ∠CDP [Alternate angles]
∴ ∆AQB ≅ ∆CPD [By SAS congruence]


(iv) ∴ AQ = CP [By CPCT]

(v) Since in APCQ, opposite sides are equal,

therefore it is a parallelogram.