Question

P and Q are points on side BC of ∆ABC such that P and Q intersect BC,
Show that ar(∆PQA) = ar(∆PAB)



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Answer 1

Answer:

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Answer 2

Given,

BP = Let height of triangle ABC with base BC and vertices " A " = h

We know Area of triangle = 1/2 × Base× Height

SO,

Area of ∆ PQA = 1/2 × PQ× h -------------- ( 1 )

And

Area of ∆ PAB = 1/2 × BP× h

given BP = PQ , SO

Area of ∆ PAB = 1/2 × PQ× h -------------- ( 2 )

And

Area of ∆ AQC = 12 × QC× h

given QC = PQ , SO

Area of ∆ AQC = 12 × PQ× h -------------- ( 3 )

SO,

From equation 1 , 2 and 3 , we get

Area of ∆ PQA = Area of ∆ PAB = Area of ∆ AQC