Question

if (mangle cbd = 130^{circ}), what is (mangle bad?)

Explanation:

Step1: Recall property of isosceles - triangle

Since $\triangle ABC$ is isosceles with $AC = BC$, $\angle CAB=\angle CBA$. Also, $\triangle ABD$ is isosceles with $AB = AD$, $\angle ABD=\angle ADB$.

Step2: Find $\angle ABC$

We know that $\angle CBD = 130^{\circ}$. Let $\angle ABC=x$. Since $\angle CBD+\angle ABC = 180^{\circ}$ (linear - pair of angles), then $x=180 - 130=50^{\circ}$.

Step3: Use angle - sum property in $\triangle ABC$

In $\triangle ABC$, since $AC = BC$, let $\angle CAB=\angle CBA = 50^{\circ}$.

Step4: Use angle - sum property in $\triangle ABD$

In $\triangle ABD$, let $\angle BAD = y$. Since $AB = AD$, $\angle ABD=\angle ADB$. And we know that $\angle ABC = 50^{\circ}$, so in $\triangle ABD$, using the angle - sum property of a triangle ($\angle BAD+\angle ABD+\angle ADB = 180^{\circ}$) and $\angle ABD=\angle ADB$, we have $y + 2\angle ABD=180^{\circ}$. Also, $\angle ABC+\angle ABD = 130^{\circ}$, so $\angle ABD = 130 - 50=80^{\circ}$. Then $y=180-2\times80 = 20^{\circ}$.

Answer:

$20^{\circ}$