In ΔABC if AL ⊥ BC and AM is the bisector of ∠A. Show that

Sum of all angles in a triangle = 180°

∠A + ∠B + ∠C = 180°

∠A = 2 ∠CAM = 2 ∠MAB {since AM is bisector of ∠A}

= 2∠CAM + ∠B + ∠C = 180°

= 2∠CAM = 180 - (∠B + ∠C)

∠AML = ∠CAM + ∠C {Exterior Angle theorem}

In Triangle ΔALM, Sum of all angles must be 180°

So, ∠LAM + ∠AML + 90 = 180

∠LAM + ∠AML = 90

∠LAM = 90 -∠AML