Basic Proportionality Theorem: A line parallel to one side of a
triangle divides the other two sides into parts of equal proportion.
Given that In triangle ABC, a line drawn parallel to BC cuts AB and AC at P
and Q respectively. See the attachment.
To Prove: AP/PB = AQ/QC
ΔABC and ΔAPQ
Since PQ is parallel to BC
∠ABC = ∠APQ
∠ACB = ∠AQP
∠BAC = ∠PAQ
Hence ΔABC ~ ΔPAQ
Since they are similar, the ratio of their sides are equal.