Question
In the figure given below, if AB is parallel to CD and CD is parallel to EF, find $\angle ACE$.
✓
Answer
Given that AB is parallel to CD and CD is parallel to EF.
$\angle E C D=180^{\circ}-130^{\circ}=50^{\circ}$
[Sum of the interior angles of same side of transversal]
$130^{\circ}+\angle E C D=180^{\circ}$
$\angle E C D=180^{\circ}-130^{\circ}=50^{\circ}$
Also, $\angle B A C=\angle A C D=70^{\circ}$ [Alternate angles]
Now, $\angle A C D=\angle E C D+\angle A C E$
$\angle A C E=\angle A C D-\angle E C D$
$\begin{array}{l}=70^{\circ}-50^{\circ} \\ =20^{\circ}\end{array}$
Therefore, $\angle A C E=20^{\circ}$.
$\angle E C D=180^{\circ}-130^{\circ}=50^{\circ}$
[Sum of the interior angles of same side of transversal]
$130^{\circ}+\angle E C D=180^{\circ}$
$\angle E C D=180^{\circ}-130^{\circ}=50^{\circ}$
Also, $\angle B A C=\angle A C D=70^{\circ}$ [Alternate angles]
Now, $\angle A C D=\angle E C D+\angle A C E$
$\angle A C E=\angle A C D-\angle E C D$
$\begin{array}{l}=70^{\circ}-50^{\circ} \\ =20^{\circ}\end{array}$
Therefore, $\angle A C E=20^{\circ}$.
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