Given : In $Δ$ ABC; BD is $⊥$ to AC, CE is $⊥$ to AB and BD = CE.
then $Δ A B C$ is isosceles.
If the above statement is true then mention answer as 1, else mention 0 if false

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Solution

Given: $△ A B C$ , $B D ⊥ A C$ and $C E ⊥ A B$ and $B D = C E$
In $△ B D C$ and $△ B C E$ ,
$∠ B D C = ∠ C E B$ (each $90^{\circ}$ )
$B C = B C$ (Common)
$B D = C E$ (Given)
Thus, $△ B D C ≅ △ C E B$ (SAS rule)
Thus, $∠ A B C = ∠ A C B$
or ABC is an Isosceles triangle.

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Given : In Δ ABC; BD is ⊥ to AC, CE is ⊥ to AB and BD = CE.then ΔABC is isosceles.If the above statement is true then mention answer as 1, else mention 0 if false

Given: △ABC, BD⊥AC and CE⊥AB and BD=CEIn △BDC and △BCE,∠BDC=∠CEB (each 90∘)BC=BC (Common)BD=CE (Given)Thus, △BDC≅△CEB (SAS rule)Thus, ∠ABC=∠ACB or ABC is an Isosceles triangle.

Given : In $Δ$ ABC; BD is $⊥$ to AC, CE is $⊥$ to AB and BD = CE.
then $Δ A B C$ is isosceles.
If the above statement is true then mention answer as 1, else mention 0 if false

Given: $△ A B C$ , $B D ⊥ A C$ and $C E ⊥ A B$ and $B D = C E$
In $△ B D C$ and $△ B C E$ ,
$∠ B D C = ∠ C E B$ (each $90^{\circ}$ )
$B C = B C$ (Common)
$B D = C E$ (Given)
Thus, $△ B D C ≅ △ C E B$ (SAS rule)
Thus, $∠ A B C = ∠ A C B$
or ABC is an Isosceles triangle.