Question

In the adjoining figure, ABCD is a parallelogram and its
diagonals AC and BD intersect at 0. P and Q are points on the
diagonal AC such that AP = CQ, Show that PBQD is a
parallelogram​

Answer 1

Given---> ABCD is a ||gm and diagonal AC and BD intersect at o and P and Q are points on the diagonal AC such that AP = CQ

To prove ---> PBQD is a ||gm.

Proof ---> In Δ APD and Δ BQC , we get,

AD = BC ( Opposite sides of ||gm )

∠ DAP = ∠ BCQ ( Alternate angles )

AP = CQ ( Given )

By SAS , Δ APD and Δ BQC are congurent

So by CPCT, DP = BC

Now , In Δ APB and Δ DQC , we get,

AB = CD ( Opposite sides of ||gm )

∠ BAP = ∠ DCQ ( Alternate angles )

AP = CQ ( Given )

So by SAS , Δ APB and congurent Δ DQC

By , CPCT , PB = DQ

Now in quadrilateral PBQD,

DP = BC ( proved )

PB = DQ ( proved )

So, PBQD is a ||gm .

Hence proved

Answer 2

Step-by-step explanation:

Given---> ABCD is a ||gm and diagonal AC and BD intersect at o and P and Q are points on the diagonal AC such that AP = CQ

To prove ---> PBQD is a ||gm.

Proof ---> In Δ APD and Δ BQC , we get,

AD = BC ( Opposite sides of ||gm )

∠ DAP = ∠ BCQ ( Alternate angles )

AP = CQ ( Given )

By SAS , Δ APD and Δ BQC are congurent

So by CPCT, DP = BC

Now , In Δ APB and Δ DQC , we get,

AB = CD ( Opposite sides of ||gm )

∠ BAP = ∠ DCQ ( Alternate angles )

AP = CQ ( Given )

So by SAS , Δ APB and congurent Δ DQC

By , CPCT , PB = DQ

Now in quadrilateral PBQD,

DP = BC ( proved )

PB = DQ ( proved )

So, PBQD is a ||gm .

Hence proved