Given: $E D = E C$
Prove: $A B + A D > B C$ .

Open in App

Solution

we know that sum of two sides of a triangle is greater than the third side.
In $△ B E C$
$B E + E C > B C$ or $B E + E D > E D / /$ since $E D = E C$
or $B D > B C - (1)$
In $△ A B D$
$A B + A D > B D$
or $A B + A D > B C$ from $(1)$

Suggest Corrections

Similar questions

E D = E C

A B + A D > B C

In the given figure; AB =BC and AD = EC. Prove that : BD = BE.

\frac{∠ B}{∠ A}

In Fig. it is given that AB = BC and AD = EC. Prove that $Δ$ ABE $≅$ $Δ$ CBD
[3 MARKS]

Join BYJU'S Learning Program