Question

9. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ
(see Fig. 8.20). Show that:
(1) ∆ APD ≡ ∆ CQB
(ii) AP=CQ
(iii) ∆AQB≡∆CPD
(iv) AQ=CP
(v) APCQ is a parallelogram
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Answer 1

Step-by-step explanation:

ANSWER

Construction: Join AC to meet BD in O.

Therefore, OB=OD and OA=OC ...(1)

(Diagonals of a parallelogram bisect each other)

But BQ=DP ...given

∴OB–BQ=OD–DP

∴OQ=OP ....(2)

Now, in □APCQ,

OA=OC ....from (1)

OQ=OP ....from (2)

∴□APCQ is a parallelogram.

In △APD and △CQB,

AD=CB ....opposite sides of a parallelogram

AP=CQ ....opposite sides of a parallelogram

DP=BQ ...given

△APD≅△CQB ...By SSS test of congruence

∴AP=CQ ...c.s.c.t.

AQ=CP ...c.s.c.t. ...(3)

In △AQB and △CPD,

AB=CD ....opposite sides of a parallelogram

AQ=CP ...from (3)

BQ=DP ...given

∴△AQB≅△CPD ....By SSS test of congruence

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Answer 2

Step-by-step explanation:

HEY MATE ....

Given ABCD is parallelogram and given BD is diagonal such that DP=BQ

⇒ Now AD∥BC [AD is parallel to BC]

BD is transversal

i.e., diagonal

In △APD & △CQBAD=CB

∠ADP=∠CBQ [Diagonals bisect the angles ]DP=BQ

∴△APD≅△CQB [SAS concurrency ]

∴AP=CQ

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