Suppose ABC is an isosceles triangle with $A B = A C$ ; BD and CE are bisectors of $∠ B$ and $∠ C$ . Prove that $B D = C E$

Question

Suppose ABC is an isosceles triangle with $A B = A C$ ; BD and CE are bisectors of $∠ B$ and $∠ C$ . Prove that $B D = C E$

Answer 1

In $△ A B C$
$A B = A C$ [Given]
$∠ A B C = ∠ A C B . . . . . . (1)$ [Angles opposite to equal sides of a trinagle are equal]
Also, $\frac{1}{2} A B = \frac{1}{2} A C$
$B E = C D . . . . . . . (2)$ [Halves of equals are equal]
Since BD and CE are two medians
Now,
In $△ B D C$ and $△ C E B$
From (1) $∠ B C D = ∠ C B E$

From (2) $B E = C D$
$B C = C B$ [Common]
$∴ △ B D C ≅ △ C E B$ [SAS Congruence rule]
$∴ B D = C E$ [By CPCT]

Answer 2

In

△ A B C

A B = A C

[Given]

∠ A B C = ∠ A C B . . . . . . (1)

[Angles opposite to equal sides of a trinagle are equal]

Also,

\frac{1}{2} A B = \frac{1}{2} A C

B E = C D . . . . . . . (2)

[Halves of equals are equal]

Since BD and CE are two medians

Now,

In

△ B D C

and

△ C E B

From (1)

∠ B C D = ∠ C B E

From (2)

B E = C D

B C = C B

[Common]

∴ △ B D C ≅ △ C E B

[SAS Congruence rule]

∴ B D = C E

[By CPCT]

Suppose ABC is an isosceles triangle with $A B = A C$ ; BD and CE are bisectors of $∠ B$ and $∠ C$ . Prove that $B D = C E$

Answer

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