In ΔPQR, angle bisector of ∠P intersects QR at M. If PQ = PR, then what is the value of ∠PMQ ?
Answer & explanation
Correct answer: option 4
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 = \({90}^\circ\).