# Find the size of the angle ACE.In a triangle ABC, <ABC=32degrees and <ACB=84degrees. Let D be a point on the side BC such that AD bisects the angle BAC. Also let E be a point on the side AB...

Find the size of the angle ACE.

In a triangle ABC, <ABC=32degrees and <ACB=84degrees. Let D be a point on the side BC such that AD bisects the angle BAC. Also let E be a point on the side AB such that BE=AC.

Find the size of the angle ACE.

*i'm not even sure where to start with this question. i really need some help to solve. what is the answer and how do i get it? if you could explain your process. any help is greatly appreciated! Thanks :)*

*print*Print*list*Cite

You should remeber that the sum of internal angles of triangle is of `180^o` . The problem provides the measures of the angles `hatB` and `hatC` , hence, you may find the measure of the angle `hatA` such that:

`hatA + hatB + hatC = 180^o`

`hatA = 180^o - (32^o + 84^o)`

`hatA = 180^o - 116^o =gt hatA = 64^o`

Notice that the problem provides the information that the lengths of the sides BE and AC are equal, hence, the congruence between triangles `hat(BAC)` and `hat(BEC)` may be proved (side-angle-side case: SAS).

Since the triangles are congruent, hence, the lengths of the sides EC and AB are equal, hence, the measures of the angles `hat(BEC)` and `hat(BAC)` are equal.

Hence, the measure of the angle `hat(BEC)` is of `84^o` => `hat(AEC) = 180^o - 84^o = 96^o` .

In triangle AEC, the sum of internal angles is of `180^o` such that:

`hat(AEC) + hat A + hat(ACE) = 180^o`

`96^o + 64^o + hat(ACE) = 180^o`

`hat(ACE) = 180^o - (96^o + 64^o)`

`hat(ACE) = 180^o - 160^o`

`hat(ACE) = 20^o`

**Hence, evaluating the measure of the angle `hat(ACE)` under the given conditions yields `hat(ACE) = 20^o` .**