Sum
In the given figure, seg AD ⊥ seg BC. seg AE is the bisector of ∠ CAB and C - E - D. Prove that
∠DAE = `1/2` (∠C - ∠B)
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Solution
In ∆ABC, since AE bisects ∠CAB, then
∠BAE = ∠CAE .......(1)
In ∆ADB,
∠ADB + ∠DAB + ∠B = 180° [Angle sum property]
⇒ 90° + ∠DAB + ∠B = 180°
⇒∠B = 90° − ∠DAB .....(2)
In ∆ADC,
∠ADC+∠DAC+∠C = 180° [Angle sum property]
⇒ 90° + ∠DAC + ∠C = 180°
⇒∠C = 90° − ∠DAC .....(3)
Subtracting (2) from (3), we get
∠C − ∠B = ∠DAC − ∠DAB
⇒ ∠C − ∠B = (∠AEC + ∠DAE) − (∠BAE − ∠DAE)
⇒ ∠C − ∠B = ∠AEC + ∠DAE − ∠BAE + ∠DAE
⇒ ∠C − ∠B = 2∠DAE [From (1)]
⇒ ∠DAE = `1/2` (∠C - ∠B)
Concept: Angle bisector theorem
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