Question

In the given figure, seg AD ⊥ seg BC. seg AE is the bisector of ∠CAB and C - E - D. Prove that ∠DAE = `1/2` (∠C - ∠B)

Answer 1

Given: seg AD ⊥ seg BC

seg AE is the bisector of ∠CAB

To prove: ∠DAE = `1/2` (∠C – ∠B)

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 C-D-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°

∴ `1/2 "∠A" + 1/2 "∠B" + 1/2 "∠C" = 90°`    ...[Dividing both side by 2]

∴ `1/2 "∠A" = 90° - 1/2 "∠C" - 1/2 "∠B"`      ...(v)

∴ ∠DAE = ∠C + `(90° 1/2 "∠C" - 1/2 "∠B") - 90°`   ...[Substituting (v) in (iv)]

∴ ∠DAE = `"∠C" - 1/2  "∠C" - 1/2  "∠B"`

= `1/2  "∠C" - 1/2  "∠B"`

 ∴ ∠DAE = `1/2`(∠C - ∠B)