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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 :

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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

Sum
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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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Chapter 11: Mid-point Theorem and Its Converse [Including Intercept Theorem] - Exercise 12 (A) [Page 151]

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Selina Concise Mathematics [English] Class 9 ICSE
Chapter 11 Mid-point Theorem and Its Converse [Including Intercept Theorem]
Exercise 12 (A) | Q 11 | Page 151

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