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Question
In ΔPQR, If PQ > PR and bisectors of ∠Q and ∠R intersect at S. Show that SQ > SR.
Solution
Given: In ΔPQR, PQ > PR and bisectors of ∠Q and ∠R intersect at S.
To prove: SQ > SR
Proof:
∠SQR = `1/2` ∠PQR ...(i) ...[Ray QS bisects ∠PQR]
∠SRQ = `1/2` ∠PRQ ...(ii) ...[Ray RS bisects ∠PRQ]
In ∆PQR,
PQ > PR ...[Given]
∴ ∠R > ∠Q ....[Angle opposite to greater side is greater.]
∴ `1/2 ("∠R") > 1/2 ("∠Q") ...["Multiplying both sides by" 1/2]`
∴ ∠SRQ > ∠SQR ...(iii) ...[From (i) and (ii)]
In ∆SQR,
∠SRQ > ∠SQR ...[From (iii)]
∴ SQ > SR ...[Side opposite to greater angle is greater]
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