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
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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.
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