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Question
In the figure, AB || RQ and BC || SQ, prove that `"PC"/"PS" = "PA"/"PR"`.
Solution
In ΔPQR, AB || RQ
∴ `"PA"/"PR" "PB"/"PQ"` ......(i)(By Basic Proportionality theorem)
In ΔPQS, BC || SQ
∴ `"PC"/"PS" = "PB"/"PQ"`......(ii)(By Basic Proportionality theorem)
From (i) and (ii)
`"PC"/"PS" = "PA"/"PR"`.
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