Q. 135.0( 3 Votes )

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

Answer :

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


triangle PDQ = triangle ADQ


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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