# Acute Angle Bisectors in a Right Triangle

The applet below illustrates problem 4 from the 1995 British Mathematical Olympiad.

Triangle ABC has a right angle at C. The internal bisectors of angles BAC and ABC meet BC and AC at P and Q respectively. The points M and N are the feet of the perpendiculars from P and Q to AB. Find angle MCN.

Solution

Triangle ABC has a right angle at C. The internal bisectors of angles BAC abd ABC meet BC and AC at P and Q respectively. The points M and N are the feet of the perpendiculars from P and Q to AB. Find angle MCN.

Since AP is an angle bisector, ∠CAP = ∠BAP. The two right triangles CAP and MAP share a hypotenuse and have equal acute angles, implying ΔCAP = ΔMAP. It follows that CP = PM, making ΔCPM isosceles. Also, AC = AM and the quadrilateral ACPM is a kite, so its diagonals are perpendicular: AP ⊥ CM.

Similarly, ∠CNQ = ∠BCN, triangle CNQ is isosceles, CQ = CN and CN ⊥ BQ.

Let CH be the altitude to AB. Then, since CH||PM||QN, ∠HCM = ∠CMP = ∠BCM and also ∠HCN = ∠CNQ = ∠ACN. It follows that

 ∠MCN = ∠HCM + ∠HCN = (∠HCM + ∠BCM + ∠HCN + ∠ACN)/2 = ∠ACB/2 = 45°.