Question

In the adjoining figure, seg AD ⊥ seg BC. Seg AE is the bisector of ∠CAB and B – E – C. Prove that ∠DAE = (1/2) (∠c – ∠B).

Answer 1

Given: seg AD ⊥ seg BC 

seg AE is the bisector of ∠CAB. 

To prove: ∠DAE = (∠C – ∠B) [∵ AD ⊥ BC] 

∴ (1/2) ∠DAE = 180° – 90° – ∠AED

∴ ∠DAE = 90° – ∠AED ….(ii) 

Proof: 

∴ ∠CAE = (1/2) ∠A ….(i) [seg AE is the bisector of ∠CAB]

In ∆DAE, 

∠DAE + ∠ADE + ∠AED = 180° [Sum of the measures of the angles of a triangle is 180°] 

∴ ∠DAE + 90° + ∠AED = 180° [∵ AD ⊥ BC] 

∴ ∠DAE = 180° – 90° – ∠AED 

∴ ∠DAE = 90° – ∠AED ….(ii) 

In ∆ACE, 

∴ ∠ACE + ∠CAE + ∠AEC = 180° [Sum of the measures of the angles of a triangle is 180°] 

∠C + -∠A + ∠AED = 180° [From (i) and CD - E] 

∴ ∠AED = 180° – ∠C – (1/2) ∠A ……(iii) 

∴ ∠DAE = 90° – 180°- ∠C + (1/2) ∠A [Substituting (iii) in (ii)] 

∴ ∠DAE = ∠C + (1/2) ∠A – 90° …..(iv) 

In ∆ABC,

∠A + ∠B + ∠C = 180°