Question

What is the x?

Answer 1

`x=32^@`

Explanation:

Triangles AEB and CEB are congruent as AE=EC, BE is common and `AhatBE=ChatBE`. So `BhatCE=58^@`.

If AD is parallel to BC then `DhatAC and BhatCE` are alternate angles and are equal. `BhatCE=DhatAC=58^@`.

Triangle DAC has `90^@ + 58^@` so `x=32^@`

Answer 2

`x=32^@`.

Explanation:

`/_ABE=/_CBE............................................."[Given]"`.

`:. BE` bisects `/_ABC" in "DeltaABC`.

Therefore, from Geometry, we know that,

`|AE|/|EC|=|AB|/|BC|...........(star)`.

But, `|AE|=|EC|......................................."[Given]"`.

`:. (star) rArr |AB|=|BC|`.

`:. /_ACB=/_BAC=58^@......(starstar)`.

Now, `AD||BC`, and `AC` is transverse

`:. /_DAC=/_ACB..."[Pair of Angles]"=58^@......[because, (starstar)]`.

Finally, in right-`DeltaADC`,

`/_ACD=180^@-{/_ADC+/_DAC}=180^@-{90^@+58^@}`.

`rArr x=/_ACD=32^@`.

Answer 3

`x=32^@`

Explanation:

`"since BE bisects AC then it is an altitude of "triangleABC`

`"and because "angleABE=angleCBE`

`"then "triangleABC" is isosceles"`

`rArrangleBCA=angleBAC=58^@tocolor(blue)"base angles"`

`rArrangleCAD=angleBCA=58^@tocolor(blue)"alternate angles"`

`"using sum of angles in a triangle "=180^@`

`rArrangleACD=x=180^@-(90+58)^@=32^@`