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