Maharashtra State BoardSSC (English Medium) 9th Standard
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In the Given Figure, Seg Ad ⊥ Seg Bc. Seg Ae is the Bisector of ∠ Cab and C - E - D. Prove that - Geometry

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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APPEARS IN

Balbharati Mathematics 2 Geometry 9th Standard Maharashtra State Board
Chapter 3 Triangles
Problem Set 3 | Q 8 | Page 50
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