Question

title=" \underline{ \underline{ \text{Question}}} : " alt=" \underline{ \underline{ \text{Question}}} : " align="absmiddle" class="latex-formula"> In the adjoining figure , AD is the bisector of $\angle$ BAC and AD $\parallel$ EC. Prove that AC = AE .
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Answer 1

Answer:

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

Answer:

See Below.

Step-by-step 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!