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[tex] \underline{ \underline{ \text{Question}}} : [/tex] In the adjoining figure , AD is the bisector of [tex] \angle[/tex] BAC and AD [tex] \parallel[/tex] EC. Prove that AC = AE .
~Thanks in advance ! ♡

tex underline underline textQuestion tex In the adjoining figure AD is the bisector of tex angletex BAC and AD tex paralleltex EC Prove that AC AE Thanks in adv class=

Respuesta :

xKelvin

Answer:

See Below.

Step-by-step explanation:

Statements:                                              Reasons:

[tex]1)\text{ } AD\text{ bisects } \angle BAC[/tex]                                  Given

[tex]2)\text{ } \angle BAD \cong \angle CAD[/tex]                                    Definition of Bisector

[tex]\displaystyle 3) \text{ } A D \parallel E C[/tex]                                               Given

[tex]\displaystyle 4)\text{ } \angle CAD\cong \angle ACE[/tex]                                    Alternate Interior Angles Theorem*

[tex]\displaystyle 5) \text{ } \angle BAD\cong \angle BEC[/tex]                                   Corresponding Angles Theorem*

[tex]6)\text{ } \angle CAD\cong \angle BEC[/tex]                                   Substitution

[tex]7)\text{ } \angle ACE\cong\angle BEC[/tex]                                    Substitution

[tex]\displaystyle 8)\text{ } AC = A E[/tex]                                              Isosceles Theorem Converse

*Please refer to the attachment below.

Let me know if you have any questions!

Ver imagen xKelvin

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.

Ver imagen Аноним

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