Question

in triangle ABC,AE is the bisector of angle BAC and AD perpendicular to BC show that angle DAE = 1\2 ( Angle C - Angle B) it's very urgent

Answer 1

In ΔABC, since AE bisects ∠A,

then ∠BAE = ∠CAE................(1)

In ΔADC,  

∠ADC+∠DAC+∠ACD = 180°   [Angle sum property]

⇒90° + ∠DAC + ∠C = 180°

⇒∠C = 90°−∠DAC...................(2)

In ΔADB,

∠ADB+∠DAB+∠ABD = 180°   [Angle sum property]

⇒90° + ∠DAB + ∠B = 180°

⇒∠B = 90°−∠DAB....................(3)

Subtracting (3) from (2), we get    

∠C − ∠B =∠DAB − ∠DAC

⇒∠C − ∠B =[∠BAE+∠DAE] − [∠CAE−∠DAE]

⇒∠C − ∠B =∠BAE+∠DAE − ∠BAE+∠DAE   [As, ∠BAE = ∠CAE ]

⇒∠C − ∠B =2∠DAE

⇒∠DAE = 1/2(∠C − ∠B)  PROVED