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Question
In the given triangle P, Q and R are the mid-points of sides AB, BC and AC respectively. Prove that triangle PQR is similar to triangle ABC.
Solution
In ∆ABC, PR || BC.
By Basic proportionality theorem,
`(AP)/(PB) = (AR)/(RC)`
Also, in ∆PAR and ∆ABC,
∠PAR = ∠BAC ...(Common)
∠APR = ∠ABC ...(Corresponding angles)
∆PAR ~ ∆BAC ...(AA similarity)
`(PR)/(BC) = (AP)/(AB)`
`(PR)/(BC) = 1/2` ...(As P is the mid-point of AB)
`(PR)/(BC) = 1/2 BC`
Similarity, `PQ = 1/2 AC`
`RQ = 1/2 AB`
Thus, `(PR)/(BC) = (PQ)/(AC) = (RQ)/(AB)`
`=>` ∆QRP ~ ∆ABC ...(SSS similarity)
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