Question

In triangle ABC, P is the mid-point of side BC. A line through P and parallel to CA meets AB at point Q; and a line through Q meets at $S$. QS parallel to BC meets median AP at point R. prove that : (i) AP = 2AR (ii) BC= 4QR

Answer 1

Given: In ∆ABC, P is the mid point of BC. PQ||CA, PQ meets AB in Q. QR||BC, QR meets AP in R.

To prove: 1. AP = 2 AR

2. BC = 4 QR

Proof:

In ∆ABC, P is the mid point of BC and PQ||AB.

∴ Q is the mid point of AB (Converse of mid-point theorem)

In ∴ ABP, Q is the mid point of AB and QR||BP.

∴ R is the mid point of AP. (Converse of mid point theorem)

⇒ AP = 2AR

In ∆ABP, Q is the mid point of AB and R is the mid point of AP.