Question

In the given figure, seg AD $\perp$ seg BC. seg AE is the bisector of $\angle$CAB and C - E - D. Prove that

$\angle$DAE = $\frac{1}{2}\left(\angle \mathrm{C}-\angle \mathrm{B}\right)$

Answer 1

In ∆ABC, since AE bisects ∠CAB, then

∠BAE = ∠CAE .......(1)
In ∆ADB,
∠ADB + ∠DAB + ∠B = 180° [Angle sum property]
⇒ 90° + ∠DAB + ∠B = 180°
⇒∠B = 90° − ∠DAB .....(2)
In ∆ADC,
∠ADC+∠DAC+∠C = 180° [Angle sum property]
⇒ 90° + ∠DAC + ∠C = 180°
⇒∠C = 90° − ∠DAC .....(3)
Subtracting (2) from (3), we get
∠C − ∠B = ∠DAC − ∠DAB
⇒ ∠C − ∠B = (∠AEC + ∠DAE) − (∠BAE − ∠DAE)
⇒ ∠C − ∠B = ∠AEC + ∠DAE − ∠BAE + ​∠DAE
​⇒ ∠C − ∠B = 2∠DAE [From (1)]
​⇒ $\angle$DAE = $\frac{1}{2}\left(\angle \mathrm{C}-\angle \mathrm{B}\right)$