Angles in Parallel Lines: Isosceles triangle

Given that CE is parallel to AD, and AB equals CB

Observe angle C and notice that the alternate angles are equal to 2X

Observe angle A and notice that the alternate angles are equal to X-10

Proceed to mark this on the drawing as follows:

Notice that angle ACE which equals 2X is supplementary to angle DAC

Supplementary angles between parallel lines equal 180 degrees.

Therefore:

$2x+DAC=180$

Let's move 2X to one side whilst maintaining the sign:

$DAC=180-2x$

We can now create an equation in order to determine the value of angle CAB:

$CAB=180-2x-(x-10)$

$CAB=180-2x-x+10$

$CAB=190-3x$

Observe triangle CAB. We can calculate angle ACB according to the law that the sum of angles in a triangle equals 180 degrees:

$ACB=180-(3x-30)-(190-3x)$

$ACB=180-3x+30-190+3x$

Let's simplify 3X:

$ACB=180+30-190$

$ACB=210-190$

$ACB=20$

Proceed to write the values that we calculated on the drawing:

Note that from the given information we know that triangle ABC is isosceles, meaning AB equals BC

Therefore the base angles are also equal, meaning:

$190-3x=20$

Let's move terms accordingly whilst maintaining the sign:

$190-20=3x$

$170=3x$

Divide both sides by 3:

$\frac{170}{3}=\frac{3x}{3}$

$x=56.67$