In ΔABC, the perpendicular drawn from A, B and C meet the opposite sides at D, E and F respectively. AD, BE and CF intersect at point P. If ∠EPD=114° and the bisectors of ∠A and ∠B meet at Q, then the measure of ∠AQB?

123°

Step 1:

∠EPD = 114°

In Quad. EPDC

∠E = ∠D = 90°

∠C = 180° - 114° = 66°

Step 2:

∠AQB = 90° + \(\frac{∠ACB}{2}\)            [Interior angle bisector theorem]

         = 90° + \(\frac{66}{2}\) = 123°

⇒∠AQB = 123°