Question

In Parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP=BQ

Show That:
(1) ∆APD=∆CQB
(2)AP=CQ
(3)∆AQB=∆CPD
(4)AQ=CP​

Answer 1

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.p.c.t.  

AQ=CP    ...c.p.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

Answer 2

Question-

In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP =BQ (see fig). show that-

1. ∆APD ≅ ∆CQB
2. AP = CQ
3. ∆AQB = ∆CPD
4. AQ = CP

Answer-

We have a parallelogram ABCD, BD is the diagonal and points P and Q are such that PD = QB

(i) Since, AD || BC and BD is a transversal.

∴ ∠ADB = ∠CBD [ ∵ Alternate interior angles are equal]

⇒ ∠ADP = ∠CBQ

Now, in ∆APD and ∆CQB, we have

AD = CB [Opposite sides of a parallelogram ABCD are equal]

PD = QB [Given]

∠ADP = ∠CBQ [Proved]

∴ ∆APD ≅ ∆CQB [By SAS congruency]

(ii) Since, ∆APD ≅ ∆CQB [Proved]

⇒ AP = CQ [By C.P.C.T.]

(iii) Since, AB || CD and BD is a transversal.

∴ ∠ABD = ∠CDB

⇒ ∠ABQ = ∠CDP

Now, in ∆AQB and ∆CPD, we have

QB = PD [Given]

∠ABQ = ∠CDP [Proved]

AB = CD [ Y Opposite sides of a parallelogram ABCD are equal]

∴ ∆AQB = ∆CPD [By SAS congruency]

(iv) Since, ∆AQB = ∆CPD [Proved]

⇒ AQ = CP [By C.P.C.T.]