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11/07/2021 11:28 am
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In Figure, 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 = \(\frac{1}{2}\) QPR.
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11/07/2021 11:30 am
Consider the ΔPQR. PRS is the exterior angle and QPR and PQR are interior angles.
PRS = QPR + PQR (According to triangle property)
PRS - PQR = QPR .........(i)
Now, consider the ΔQRT,
TRS = TQR + QTR
QTR = TRS - TQR
We know that QT and RT bisect PQR and PRS respectively.
PRS = 2 TRS and PQR = 2TQR
Now, QTR = \(\frac{1}{2}\) PRS – \(\frac{1}{2}\)PQR
Or, QTR = \(\frac{1}{2}\) (PRS -PQR)
From (i) we know that PRS -PQR = QPR
QTR = \(\frac{1}{2}\) QPR (hence proved).