In the given diagram, AD = AB and AE bisects $∠ A$ , then
BE = DE.
If the above statement is true then mention answer as 1, else mention 0 if false.

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Solution

In $Δ$ ABC, AE bisect $∠$ A and cut line BC at point E,
as AB = AD(given), line AE is common.
$Δ$ ABE or $Δ$ ADE are similar triangle,

Then BE = DE
So the above statement is true then mention answer as 1.

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Δ A B D ≅ Δ C B D

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Given : ED = EC

∠ B A C

In the given figure, AD = AE and $A D^{2} = B D \times E C$ .

Hence , $△ A B D$ $\sim$ $△ C A E$ .

If the above statement is true then mention answer as 1, else mention 0 if false

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