Question

If AD is the bisector of ∠BAC in ∆ABC, AL ⊥ BC, then prove that ∠DAL = ½( ∠B –∠C ).​

Answer 1

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sanjanaregulagedda

sanjanaregulagedda

06.09.2020

Math

Secondary School

answered

If AD is the bisector of ∠BAC in ∆ABC, AL ⊥ BC, then prove that ∠DAL = ½( ∠B –∠C ).

Answer 2

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$\huge{\underline{\underline{\sf{\pink{</p><p>Question:-}}}}}$

If AD is the bisector of ∠BAC in ∆ABC, AL ⊥ BC, then prove that ∠DAL = ½( ∠B –∠C ).

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$\huge{\underline{\underline{\sf{\orange{</p><p>Solution:-}}}}}$

Since AE is the bisector of ∠BAC,we have

$\angle \: BAE = \angle \: CAE \: \: \: \: \: \: ....(i)$

In ∆ABC,we have ∠ADB= 90°

$\therefore \: \angle \: ABD + \angle \: BAD = 90\degree$

$⇒ \angle \: ABD = (90\degree - \angle \: BAD)$

$⇒ \angle \: B \: = (90\degree - \angle \: BAD). \: \: \: \: \: \: ....(ii)$

In ∆ADC,we have ∠ADC = 90°

$\therefore \: \angle \: CAD + \angle \: ACD = 90\degree$

$⇒ \angle \: ACD = (90\degree - \angle \: CAD)$

$⇒ \angle \: C = (90\degree - \angle \: CAD). \: \: \: \: \:...(iii)$

On subtracting (iii) from (ii),we get

$(\angle \: B - \angle \: C) = (90\degree - \angle \: BAD) - (90\degree - \angle \: CAD) \\ = \angle \: CAD - \angle \: BAD \\ = (\angle \: CAE + \angle \: DAE) - (\angle \: BAE - \angle \: DAE) \\ = 2\angle \: DAE \: \: \: \: \: [using \: (i)].$

$\therefore \: DAE = \frac{1}{2} (\angle \: B - \angle \: c).$

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