Question

In Parallelogram ABCD, two points P and Q are
taken on diagonal BD such that DP BQ
Show that : i) AAPD = ACQB
ii) AP = CQ
iii) AAQB = ACPD​

Answer 1

Given: ABCD is a parallelogram where DP = BQ

i) To prove: ∆APD ≅ ∆CQB

AD ∥ BC and transversal BD (Opposite sides of parallelogram are parallel)

∠ADP = ∠CBQ (Alternate angles) ......(1)

In ∆APD and ∆CQB,

AD = CB (Opposite sides of parallelogram are equal)

∠ADP = ∠CBQ (From (1))

DP = BQ (Given)

∴ ∆APD ≅ ∆CQB (By SAS condition of congruency)

ii) AP = CQ (As ∆APD ≅ ∆CQB are congruent)

iii) To prove: ∆AQB ≅ ∆CPD

Since, AB ∥ DC and transversal BD (Opposite sides of parallelogram are parallel)

∠ABQ = ∠CDP (Alternate angles) ......(2)

In ∆AQB and ∆CPD,

AB = CD (Opposite sides of parallelogram are equal)

∠ABQ = ∠CDP (From (2))

DP = BQ (Given)

∴ ∆AQB ≅ ∆CPD (By SAS condition of congruency)

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

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]