Question:
In the following figure, \(\frac{QR}{QS}=\frac{QT}{PR}\) and \(\angle 1 =\angle 2\)
Show that ΔPQR∼ΔTQR
In the following figure, \(\frac{QR}{QS}=\frac{QT}{PR}\) and \(\angle 1 =\angle 2\)
Show that ΔPQR∼ΔTQR
Show that ΔPQR∼ΔTQR
Updated On: Nov 2, 2023
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Solution and Explanation
Given: \(\frac{QR}{QS}=\frac{QT}{PR}\) and \(\angle 1 =\angle 2\)
To Prove: ΔPQR∼ΔTQR
Proof: In ∆PQR,
\(\angle\)PQR = \(\angle\)PRQ
∴ PQ = PR ………………(i)
Using (i) we obtain
\(\frac{QR}{QS}=\frac{QT}{QP}\).............(ii)
In ΔPQS and ΔTQR,
\(\frac{QR}{QS}=\frac{QT}{QP}\) [using (ii)]
\(\angle\)Q=\(\angle\)Q
∴ ΔPQS∼ΔTQR
Hence Proved
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