Question

IN FIGURE ANGLE BED =ANGLE CED AND AD IS THE BISECTOR OF
ANGLE BAC PROVE THAT BE = CE

Answer 1

Given:

∠BED = ∠CED

AD is the bisector of ∠BAC

To prove:

BE = CE

Solution:

We have,

AD is the bisector of ∠BAC

∴ ∠BAE = ∠CAE ..... (i)

From the figure attached below, we can say,

∠BED + ∠BEA = 180° ..... (ii) ..... [Linear Pair]

and

∠CED + ∠CEA = 180° ..... (iii) ..... [Linear Pair]

From (ii) & (iii), we get

∠BED + ∠BEA = ∠CED + ∠CEA

∵ ∠BED = ∠CED .... (given)

⇒  ∠BEA = ∠CEA ...... (iv)

Now,

In Δ ABE and Δ ACE, we have

∠BAE = ∠CAE ..... [From (i)]

EA = EA ....... [common side]

∠BEA = ∠CEA ...... [From (iv)]

∴ Δ ABE ≅ Δ ACE ...... [by ASA congruency]

We know that ⇒ C.P.C.T. ⇒ this theorem states if two triangles are congruent to each other then the corresponding angles and the sides of the triangles are also congruent to each other.

Here we got, Δ ABE ≅ Δ ACE

∴ By C.P.C.T. → $\boxed{\bold{BE = CE}}$

Hence Proved

$\star$ Note: Figure is given as an attachment below $\star$

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Answer 2

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