In ΔPQR, angle bisector of ∠P intersects QR at M. If PQ = PR, then what is the value of ∠PMQ ?

90 degree

In \(\Delta \)PQR, PQ = PR

= \(\angle\)Q = \(\angle\)R = b

PM is the angle bisector of \(\angle\)P.

\(\angle\)QPM = \(\angle\)RPM = a

In \(\Delta \)PQR, apply angle sum property

\(\angle\)P + \(\angle\)Q + \(\angle\)R = 180

\(\angle\)P + b + b = 180

\(\angle\)P = 180 - 2b

\(\angle\)QPM = \(\frac{(180\; -\; 2b)}{2}\) = 90 - b

In \(\Delta \)PQM

Let \(\angle\)PMQ = \(\theta \)

\(\angle\)QPM + \(\angle\)Q + \(\angle\)M = 180

= 90 - b + b + \(\theta \) = 180

= \(\theta \) = 180 - 90

= \(\theta \) = \({90}^\circ\)

Therefore, \(\angle\)PMQ = \({90}^\circ\).