Question

In parallelogram ABCD, two points P and Q are taken on diagonal BD such
that DP = BQ. Show that:
(i) ΔAPD ≅ ΔCQB
(ii) AP = CQ
(iii) ΔAQB ≅ ΔCPD
(iv) AQ = CP

Answer 1

Step-by-step explanation:

ABCD is a parallelogram

DP = BQ

(i) ∆APD ≅ ∆CQB

As ABCD is a parallelogram

∠ADB = ∠CBD [Alternate interior angles are equal]……………….(1)

∠ABD = ∠CDB [Alternate interior angles are equal]…………………(2)

Now, in ∆APD and ∆CQB, we have

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

PD = QB [Given]

∠ADP = ∠CBQ

Hence, ∆APD ≅ ∆CQB [By SAS congruency]

(ii) AP = CQ

AP = CQ by CPCT as ΔAPD ≅ ΔCQB.

(iii) ∆AQB ≅ ∆CPD

in ∆AQB and ∆CPD, we have

QB = PD [Given]

∠ABQ = ∠CDP [Proved]

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

Hence, ∆AQB ≅ ∆CPD [By SAS congruency]

(iv) AQ = CP

As, ∆AQB ≅ ∆CPD [Proved]

AQ = CP [By C.P.C.T.] …………………………..(4)