Lines ac and bd are parallel to CB bisector of angle ACD. Prove that triangle BCD is isosceles.

Let’s draw straight lines AC and BD. Draw the bisector CB from the ACD angle. We get a triangle CBD. How can one prove that he is isosceles?
Of course, on two equal sides it will not work here. Therefore, we will use the fact that in any p / w triangle the angles at the base are equal. Thus, if we prove that two angles in the CBD triangle are equal, then we can safely say that it is isosceles.
If you look closely at the drawing, we can say that AC and BD are two parallel straight lines, and continue our bisector CB and take that it is a secant for AC and BD, then:
The bisector divides the angle in half. Hence, BCD = ACB. If SV is a secant, then the angle ACB is internal crosswise with the angle DBC. And for parallel straight lines, the inner angles lying crosswise are equal. This means that if the angle ACB = CBD, then it is equal to BCD.
We get that in the triangle DCB two angles are equal. This means that it is isosceles.