Q13 of 31 Page 232

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


∴ 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


More from this chapter

All 31 →
11

AD and BE are the medians of triangle ABC. Let’s prove that triangle ACD = triangle BCE.

 12

A line parallel to BC of triangle ABC intersects sides AB and AC at the points P and Q respectively. CQ and BQ intersect each other at the point X. Let’s prove that.

(i) ΔBPQ = triangle CPQ


(ii) ΔBCP = triangle BCQ


(iii) ΔACP = triangle ABQ


(iv) ΔBXP = triangle CXQ

 14

In triangle ABC of which AB = AC; perpendiculars through the points B and C on sides AC and AB intersect sides AC and AB at the points E and F. Let’s prove that, FE || BC.

 15

In triangle ABC ∠ABC = ∠ACB; bisectors of an angle ∠ABC and ∠ACB intersect the side AC and AB at the points E and F respectively. Let’s prove that, FE || BC.