an equilateral triangle and an isosceles triangle share a common side. what is the measure of $ \angle abc $?
the measure of $ \angle abc $ is $ \square ^ { \circ } $.
(simplify your answer.)

Question

Accepted Answer

# Explanation:
## Step1: Analyze triangle $ABD$
Since $AB = BD$ (given by the marks on the sides), $\triangle ABD$ is isosceles. In $\triangle ABD$, $\angle A=70^{\circ}$, and using the isosceles - triangle property ($\angle A=\angle ADB = 70^{\circ}$), then $\angle ABD=180^{\circ}-\angle A-\angle ADB$.
$$
\begin{align*}
\angle ABD&=180^{\circ}-70^{\circ}-70^{\circ}\
&=40^{\circ}
\end{align*}
$$
## Step2: Analyze triangle $BDC$
Since $BD = BC$ (given by the marks on the sides), $\triangle BDC$ is isosceles. Also, assume $AB = BD = BC$ (equilateral - triangle and isosceles - triangle share a common side). $\triangle BDC$ is equilateral (because $BD = BC$ and if we assume the side - length equality from the common - side property in a proper geometric - figure context, all sides $BD = BC=DC$). So $\angle DBC = 60^{\circ}$ (angles in an equilateral triangle are $60^{\circ}$).
## Step3: Calculate $\angle ABC$
$\angle ABC=\angle ABD+\angle DBC$
$$
\begin{align*}
\angle ABC&=40^{\circ}+60^{\circ}\
&=100^{\circ}
\end{align*}
$$

# Answer:
$100$

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QUESTION IMAGE

Question

Explanation:

Step1: Analyze triangle \(ABD\)

Step2: Analyze triangle \(BDC\)

Step3: Calculate \(\angle ABC\)

Answer: