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प्रश्न
In the given figure, if a circle touches the side QR of ΔPQR at S and extended sides PQ and PR at M and N, respectively, then prove that PM = `1/2` (PQ + QR + PR)
उत्तर
Given: A circle is touching a side QR of ΔPQR at point S.
PQ and PR are produced at M and N respectively.
To prove: PM = `1/2` (PQ + QR + PR)
Proof: PM = PN ...(i) (Tangents drawn from an external point P to a circle are equal)
QM = QS ...(ii) (Tangents drawn from an external point Q to a circle are equal)
RS = RN ...(iii) (Tangents drawn from an external point R to a circle are equal)
Now, 2PM = PM + PM
= PM + PN ...[From equation (i)]
= (PQ + QM) + (PR + RN)
= PQ + QS + PR + RS ...[From equations (i) and (ii)]
= PQ + (QS + SR) + PR
= PQ + QR + PR
∴ PM = `1/2` (PQ + QR + PR)
Hence proved.
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