if the bisector of an angle of a triangle bisector the opposite side to that the triangle is isosceles
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In a ∆ABC,Consider AD be the bisector of ∠A then BD = CD.
To prove that ∆ABC is an isosceles triangle i.e. AB = AC. Draw a line from C i.e CE parallel AD .
BA is extended then they meet at E.
Given that ∠BAD = ∠CAD .............. (i)
CE || AD
∴ ∠BAD = ∠AEC (Corresponding angles) ................ (ii)
And ∠CAD = ∠ACE (Alternate interior angles) .................. (iii)
From (i), (ii) and (iii) we obtain
∠ACE = ∠AEC
In ∆ACE, ∠ACE = ∠AEC
∴ AE = AC (Sides opposite to angles are equal) ................ (iv)
In a ∆BEC, AD||CE and D is the mid-point of BC by converse of mid-point theorem A is the mid-point of BE.
∴ AB = AE
⇒ AB = AC [equ (iv)]
In a ∆ABC, AB = AC
∴ ∆ABC is an isosceles triangle.
To prove that ∆ABC is an isosceles triangle i.e. AB = AC. Draw a line from C i.e CE parallel AD .
BA is extended then they meet at E.
Given that ∠BAD = ∠CAD .............. (i)
CE || AD
∴ ∠BAD = ∠AEC (Corresponding angles) ................ (ii)
And ∠CAD = ∠ACE (Alternate interior angles) .................. (iii)
From (i), (ii) and (iii) we obtain
∠ACE = ∠AEC
In ∆ACE, ∠ACE = ∠AEC
∴ AE = AC (Sides opposite to angles are equal) ................ (iv)
In a ∆BEC, AD||CE and D is the mid-point of BC by converse of mid-point theorem A is the mid-point of BE.
∴ AB = AE
⇒ AB = AC [equ (iv)]
In a ∆ABC, AB = AC
∴ ∆ABC is an isosceles triangle.
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