in triangle ABC , AM is the bisector of angle A , AN is perpendicular to BC.    proove that Angle MAN=1/2(angleB -angle C)

Given:

In ΔABC

AM is the bisector of ∠A

⇒∠BAM = ∠MAC  .......(i)

and AN ⊥ BC

In ΔABN

∠ABN + ∠BAN + ∠ANB = 180°

⇒∠B = 180° – ∠ANB – ∠NAB

= 180° – 90° – ∠NAB

= 90° – ∠NAB  ......(2)

Similarly ΔANC

∠ANC + ∠ACN + ∠NAC = 180°

⇒∠C = 90° – ∠NAC  ......(3)

Subtracting (3) from (2) we get

∠B – ∠C = 90° – ∠NAB – (90° – ∠NAC)

= ∠NAC – ∠NAB

= (∠NAM + ∠MAC) – (∠BAM – ∠MAN)

= 2∠ MAN