Q. 125.0( 3 Votes )

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

Answer :

Given.


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


Formula used.


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



In triangle PQB and triangle PQC


As both triangles lies on same base PQ


And both are between parallel lines as PQ || BC


triangle PQB = triangle PQC


In triangle BCP and triangle BCQ


As both triangles lies on same base BC


And both are between parallel lines as PQ || BC


triangle BCP = triangle BCQ


In triangle ACP and triangle ABX


As triangle PQB = triangle PQC proved above


Add triangle APQ in both triangles


triangle PQB + triangle APQ = triangle PQC + triangle APQ


triangle ACP = triangle ABX


In triangle BXP and triangle CXQ


As triangle PQB = triangle PQC proved above


Subtract triangle PXQ in both triangles


triangle PQB – triangle PXQ = triangle PQC – triangle PXQ


triangle BXP = triangle CXQ


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