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In figure, if angle1=angle2 and DeltaNSQ...

In figure, if `angle1=angle2` and `DeltaNSQ=DeltaMTR`, then prove that `DeltaPTS~DeltaPRQ`.

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Given: In `trianglePQR` points S in on PQ and T is on PR.
such that ` angle1= angle2`
and ` triangleNSQ cong triangleMTR`
To prove : ` trianglePTS ~ trianglePRQ`
Proof : ` triangleNSQ cong triangleMTR` ( given)
SQ = TR ( CPCT) ... (1)
`angle1= angle2`
PT = PS ( sides opposite to equal angles in `trianglePTS`) ...(2)
` Rightarrow " " (PT)/(TR)= (PS)/(SQ)` [ from (1), (2)]
ST||QR ( by converse of BPT)
Now in ,` trianglePTS and trianglePRQ` ,we have
ST||QR ( prove above)
`angle1=angle3` (corresponding angles)
`angle2 = angle4` ( corresponding angles)
`trianglePTS ~ trianglePRQ` ( by AA similartiy criterion) Hence proved.
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