| In the adjoining figure, \[\frac{PS}{SQ}=\frac{PT}{TR}\] and \[\angle PST=\angle PRQ\]. Then, \[\Delta PQR\] is an |
 |
A) equilateral triangle
B) right angle triangle
C) isosceles triangle
D) Cannot say
Correct Answer: C
Solution :
| Given, \[\frac{PS}{SQ}=\frac{PT}{TR}\] and \[\angle PST=\angle PRQ\] |
| Since, \[\frac{PS}{SQ}=\frac{PT}{TR}\] |
| \[\therefore \,\,\,\,ST\,\,|\,\,|\,\,QR\] |
| [by converse of basic proportionality theorem] |
| Then, \[\angle PST\,\,=\angle PQR\] |
| [corresponding angles] ...(i) |
| Also, \[\angle PST\,\,=\angle PRQ\] [given] ...(ii) |
| From Eqs. (i) and (ii), we get |
| \[\angle PRQ=\angle PQR\] |
| \[\Rightarrow \,\,\,\,PQ=PR\] |
| [since, sides opposite to equal angles of a triangle are also equal] |
| Hence, \[\Delta PQR\] is an isosceles triangle. |
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