In the figure, Angle DAB=20 degrees and AB = AD = CD. Find the value of x and y. x being angle ADB, y being angle BDC.
All help is appreciated. DCB is 20 degrees right?
You need to attach the figure or we cannot tell what DCB is.
Since AB = AD, $\displaystyle \triangle ABD$ is isosceles so the base angles are the same.
Also, the angles of a triangle add up to $\displaystyle 180^o$
Now see if you can continue.
Hello, Rocher!
We still need more specifics.
Do we have a parallelogram?
According to your description, CD can be anywhere . . .In the figure: .$\displaystyle \angle DAB = 20^o,\;AB = AD = CD$
Find: .$\displaystyle x = \angle ADB,\:y = \angle BDC.$Code:D * * x * * * * y * * * * * * * C * 20° * * * * * * * * * * * * A B
This diagram is messed up somehow. Every statement you made above has something wrong with it!
Angle x is ABD, not ADB. Angle y is CAD, not BDC. You have AB = AD = AC, not AB = AD = CD. DCB is a straight line segment, not an angle. And though this is not a critical point to make, angle DAB doesn't look anywhere near 20 degrees on your diagram.
-Dan