# 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

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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