Question

in the given figure AD is the bisector of angle BAC and AM is perpendicular to BC .show that angle DAM=1/2(angle C-angle D)

Answer 1

n ∆ABC, since AE bisects ∠A, then

∠BAE = ∠CAE .......(1)

In ∆ADC,

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

⇒90° + ∠DAC + ∠C = 180°⇒∠C = 90°−∠DAC .....(2)

In ∆ADB,

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

⇒90° + ∠DAB + ∠B = 180°

⇒∠B = 90°−∠DAB .....(3)

Subtracting (3) from (2), we get

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

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

⇒∠C − ∠B =2∠DAE

⇒∠DAE = 1/2(∠C − ∠B)

hope it's help u (✽ ﾟдﾟ ✽)