Answers

2015-10-02T09:57:23+05:30
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

Proof:
In 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.

 \frac{AB}{AP} = \frac{AC}{AQ}\\ \\ \Rightarrow \frac{AB}{AP}-1 = \frac{AC}{AQ}-1\\ \\ \Rightarrow \frac{AB-AP}{AP} = \frac{AC-AQ}{AQ}\\ \\ \Rightarrow \frac{PB}{AP} = \frac{QC}{AQ}\\ \\ \Rightarrow \frac{AP}{PB} = \frac{AQ}{AC}

Hence, Proved!
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