Question

quadrilateral ABCD ia a parelelogram p is any point on side bc find two pair of triangle with equal area ​

Answer 1

Answer:

Join AC.

As we know that area of triangles on the same base and between the same parallels lines are equal

Therefore , ar(APC)=ar(BPC)... (1)

Now In quadrilateral ACQD, we have

AD=CQ

and , AD∥CQ [Given]

Therefore, this quadrilateral ADQC with one pair of opposite sides is equal and parallel is parallelogram.

Therefore ADQC is a parallelogram.

⇒AP=PQ and CP=DP

[Since diagonals of a|| gm bisect each other]

In Δs APC and DPQ we have

AP=PQ [Proved above]

∠APC=∠DPQ [Vertically opp. ∠s]

and ,

PC=PD [Proved above]

Therefore by SAS criterion of congruence,

ΔAPC≅ΔDPQ

⇒ar(APC)=ar(DPQ) ... (2)

[Since congruent Δs have equal area]

Therefore ar(BPC)=ar(DPQ) [From (1)]

Hence , ar(BPC)=ar(DPQ)

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Answer 2

Answer:

please refer to the image..!!!!

Step-by-step explanation:

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