Angles in Parallel Lines: Isosceles triangle

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

Let's look at angle C and notice that the alternate angles are equal to 2X

Let's look at angle A and notice that the alternate angles are equal to X-10

Let's mark this on the drawing as follows:

Now let's notice that angle ACE which equals 2X is supplementary to angle DAC

Meaning supplementary angles between parallel lines equal 180 degrees.

Therefore:

$2x+DAC=180$

Let's move 2X to one side and keep the appropriate sign:

$DAC=180-2x$

Now we can create an equation to find the value of angle CAB:

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

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

$CAB=190-3x$

Now let's look at 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:

$ACB=180+30-190$

$ACB=210-190$

$ACB=20$

Let's write the values 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 and keep the appropriate sign:

$190-20=3x$

$170=3x$

Let's divide both sides by 3:

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

$x=56.67$