Question

in an equilateral triangle abc, if the bisector of angle b and angle c meet at a point D, then prove that AD+ BD> AB​

Answer 1

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Given A △ABC in which the bisectors of ∠B and ∠C meet the sides AC and AB at D and E respectively.

To prove AB=AC

Construction Join DE

Proof In △ABC, BD is the bisector of ∠B.

∴    

BC

AB

​  

=  

DC

AD

​  

...........(i)

In △ABC, CE is the bisector of ∠C.

∴    

BC

AC

​  

=  

BE

AE

​  

.......(ii)

Now, DE∣∣BC

⇒    

BE

AE

​  

=  

DC

AD

​  

            [By Thale's Theorem]......(iii)

From (iii), we find the RHS of (i) and (ii) are equal. Therefore, their LHS are also equal i.e.

       

BC

AB

​  

=  

BC

AC

​  

 

⇒   AB=AC

Hence,  △ABC is isosceles.