Question

in a triangle ABC, AE is the bisector of angle BAC. AD is perpendicular to BC. show that angle DAE=1/2(angle C- angle B)

Answer 1

RTP :- angle DAE=1/2(angle C- angle 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

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Answer 2

In ∆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)