Angles in Parallel Lines: Isosceles triangle

To demonstrate that triangle AED is isosceles, we must prove that its hypotenuses are equal or that the opposite angles to them are equal.

Given that angles ABC and ACB are equal (since they are equal opposite bisectors),

And since ED is parallel to BC, the angles ABC and ACB alternate and are equal to angles ADE and AED (alternate and equal angles between parallel lines)

Opposite angles ADE and AED are respectively sides AD and AE, and therefore are also equal (opposite equal angles, the legs of triangle AED are also equal)

Therefore, triangle ADE is isosceles.

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:

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

$x=56.67$