In triangle ABC angle B angle C , bisector of angle A meets BC at O and AE - Maths - Lines and Angles

Question

In triangle ABC angle B > angle C , bisector of angle A meets BC
at O and AE perpendicular to BC Prove that:-2 angleDAE=(angleB -
angleC)

Akash Kumar · Accepted Answer

Since, AD bisects ∠BAC, so∠BAD = ∠DAC   
1In ∆AEB,    ∠BAE + ∠ABE + ∠AEB = 180° Angle sum property⇒∠BAE + ∠B  + 90° = 180°⇒∠BAE + ∠B = 180° - 90°⇒∠BAE + ∠B = 90°⇒∠B = 90° - ∠BAE   
2In ∆AEC,∠ACE + ∠EAC + ∠AEC = 180°   Angle sum property⇒∠C + ∠EAC + 90° = 180°⇒∠C + ∠EAC = 90°⇒∠C = 90° - ∠EAC     
3subtracting 3 from 2, we get     ∠B - ∠C = 90° - ∠BAE - 90° - ∠EAC⇒∠B - ∠C  = ∠EAC - ∠BAE⇒∠B - ∠C  = ∠EAD + ∠DAC - ∠BAD - ∠EAD⇒∠B - ∠C = 2∠EAD + ∠DAC - ∠BAD⇒∠B - ∠C = 2∠EAD    As, ∠BAD = ∠DAC⇒2∠DAE =∠B - ∠C

In triangle ABC angle B > angle C , bisector of angle A meets BC at O and AE perpendicular to BC . Prove that:-2 angleDAE=(angleB - angleC)