Question

If bisector of an angle of a triangle also bisects the opposite side,prove that the triangle is isosceles.​

Answer 1

If the bisector of an angle of a triangle also bisects the opposite side then prove that the triangle is isosceles?

Let ABC. is a triangle in which AD. is a bisector of angle BAC which meets the. BC. on D , such that BD= CD.

To prove:- Triangle ABC is an isosceles triangle.

Proof:- In triangle BAD. and triangle CAD.

angle BAD. = angle. CAD. . (given )

side. BD. = side CD. (given)

AD. is common.

∆ BAD. is congruence. ∆ CAD

Therefore side AB. = side. AC. or. ∆ ABC is an isosceles triangle. Proved.

Hope it help u

Answer 2

Sol: 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.

PLEASE FOLLOW ME......