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 and parallel to BC meets median AP at point R.
Prove that : (i) AP = 2AR
                   (ii) BC = 4QR

Answer 1

The required figure is shown below

From the figure, it is seen that P is the midpoint of BC and PQ || AC and QR || BC
Therefore Q is the midpoint of AB and R is the midpoint of AP
(i) Therefore AP=2AR
(ii) Here we increase QR so that it cuts AC at S as shown in the figure.
(iii) From triangle PQR and triangle ARS
∠PQR = ∠ARS                   ...( Opposite angle )
PR = AR
PQ = AS                            ...[ PQ = AS = `1/2`AC ]
ΔPQR ≅ ΔARS                   ...( SAS Postulate )
Therefore QR = RS
Now,
BC = 2QS
BC = 2 x 2QR
BC = 4QR 
Hence proved.