Question

best answer will be marked brainliest....​

Answer 1

Answer:

Hey!!!

Given that, AB=BP=8cm, AD=5cm, AC=10cm

(i) In ∆APQ, BC || PQ (Given)

B is the mid point of AP

=> By mid point theorem in a ∆,

C is also the mid point of AQ.

Hence, proved.

(ii) In Quadrilateral BCQP, BC || AD {Given}

DC || AB {Given}

Thus, BCQP is a || gm, as paris of opposite sides are parallel.

Thus, Perimeter(BCQP)= 2×(AB+AD)

=2×(8+5)=26cm.

I hope it helps you ^_^

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