Question

give the full solution and no spam the correct would be the brainliest.

Answer 1

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Give solution of the question shown in picture.

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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)

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