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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

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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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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).

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