Question

In triangleABC, 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
and parallel to BC meets median AP at point
R. Prove that: (i) AP = 2AR (ii) BC = 4QR with diagram​

Answer 1

Answer:



Step-by-step explanation:

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

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

QR = 1/2 BP mid point thorem

QR = 1/4 BC