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In figure, if `angle1=angle2` and `DeltaNSQ=DeltaMTR`, then prove that `DeltaPTS~DeltaPRQ`. |
Answer» Given `DeltaNSQcongDeltaMTRandangle1=angle2` to prove `DeltaPTS~DeltaPRQ` Proof Since, `DeltaNSQcongDeltaMTR` So, SQ=TR…..(i) Also, `angle1=angle2rArr`PT=PS..(ii) [since, sides opposite to equal angles are also equal] From Eqs. (i) and (ii) `(PS)/(SQ)=(PT)/(TR)` `rArr STabs()QR` [by convense of basic proportionality theorem] `thereforeangle1=anglePQR` and `angle2=anglePRQ` in `DeltaPTS and Delta PRQ` [common angles] `angleP=angleP` `angle1=anglePQR` `angle2=anglePRQ` `thereforeDeltaPTS~DeltaPRQ` [by AAA similarity criterion] Hence proved |
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