Question

solve this problem guys​

Answer 1

Step-by-step explanation:

In ∆ABC, since AE bisects ∠A, then

∠CAE = ∠BAE------(1)

In ∆ADB,

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

⇒90° + ∠DAB + ∠B = 180°⇒∠B = 90°−∠DAB ------(2)

In ∆ADC,

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

90° + ∠DAC + ∠C = 180°

∠C = 90°−∠DAC ------(3)

Subtracting (3) from (2), we get

∠B − ∠C =∠DAC − ∠DAB⇒∠B − ∠C =[∠CAE+∠DAE] − [∠BAE−∠DAE]

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

∠B − ∠C =2∠DAE

∠DAE = 1/2(∠B − ∠C)