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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
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Solution
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.
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