Sir math me ex 2.3 ka …
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Gaurav Seth 3 years, 11 months ago
6. In Fig. 6.44, the side QR of ΔPQR is produced to a point S. If the bisectors of PQR and PRS meet at point T, then prove that QTR = ½ QPR.
Solution:
Consider the ΔPQR. PRS is the exterior angle and QPR and PQR are interior angles.
So, PRS = QPR+PQR (According to triangle property)
Or, PRS -PQR = QPR ———–(i)
Now, consider the ΔQRT,
TRS = TQR+QTR
Or, QTR = TRS-TQR
We know that QT and RT bisect PQR and PRS respectively.
So, PRS = 2 TRS and PQR = 2TQR
Now, QTR = ½ PRS – ½PQR
Or, QTR = ½ (PRS -PQR)
From (i) we know that PRS -PQR = QPR
So, QTR = ½ QPR (hence proved).
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