Question

`\underline{ \underline{ \text{Question}}} :` In the adjoining figure , AD is the bisector of
`\angle` BAC and AD
`\parallel` EC. Prove that AC = AE .

~Thanks in advance ! ♡

Answer 1

In the given figure AD is internal bisector of angle A and CE is parallel to DA. If CE meets BA produced at E.

to prove :AC=AE

<BAC + <EAC = 180° ( Straight line)

In triangle ACE

<ACE + <AEC + <EAC = 180° (sum ofangles of Triangle)

Equating both

<BAC + <EAC = <ACE + <AEC + <EAC

=<BAC = <ACE + <AEC.......... Eq 1

<BAD = (1/2) <BAC

<BAD = <AEC so (AD || CE)

we get

<AEC = (1/2) <BAC

putting this in eq 1

=<BAC = <ACE + <AEC Eq1

<BAD = (1/2) <BAC

<BAD = <AEC (AD || CE)

=<AEC = (1/2) <BAC

<AEC=<ACE=1/2 <BAC

: AC = AE

hence proved.

Answer 2

See Below.

Explanation:

Statements: Reasons:

`1)\text{ } AD\text{ bisects } \angle BAC` Given

`2)\text{ } \angle BAD \cong \angle CAD` Definition of Bisector

`\displaystyle 3) \text{ } A D \parallel E C` Given

`\displaystyle 4)\text{ } \angle CAD\cong \angle ACE` Alternate Interior Angles Theorem*

`\displaystyle 5) \text{ } \angle BAD\cong \angle BEC` Corresponding Angles Theorem*

`6)\text{ } \angle CAD\cong \angle BEC` Substitution

`7)\text{ } \angle ACE\cong\angle BEC` Substitution

`\displaystyle 8)\text{ } AC = A E` Isosceles Theorem Converse

*Please refer to the attachment below.

Let me know if you have any questions!