# D is the midpoint of BC of triangle ABC and P is any a point on BC. Join P, A-Through the point D a straight line parallel to line segment PA meets AB at point Q. Let’s prove that,(i) ΔADQ = triangle PDQ(ii) ΔBPQ = ΔABC

Given.

D is the midpoint of BC of triangle ABC

From point D a straight line parallel to line segment PA meets AB at point Q

Formula used.

If 2 triangles are on same base And lies between 2 parallel lines then both triangles are equal.

Join DA In triangle PDQ and triangle ADQ

As both triangles lies on same base DQ

And both are between parallel lines as PA || DQ

As AD is median of triangle ABC

triangle ADC = triangle ABD = × triangle ABC

triangle ABD = triangle BDQ + triangle ADQ

triangle ABD = triangle BDQ + triangle PDQ proved above × triangle ABC = triangle BPQ

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