Question

In tri. ABC, angle B > angle C. If AM is the bisector of angle ABC and AN perpendicular to BC. prove that angle MAN =½(angle B - angle C) ​

Answer 1

Step-by-step explanation:

B)

In ΔABC, since AE bisects ∠A,

then ∠BAE = ∠CAE................(1)

In ΔADC,

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

⇒90° + ∠DAC + ∠C = 180°

⇒∠C = 90°−∠DAC...................(2)

In ΔADB,

∠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) PROVED