Question

In the diagram $\overleftrightarrow{\mathit{\text{AB}}}$bisects $\mathit{\angle }\mathit{\text{DHF}}$, $\overleftrightarrow{\mathit{\text{AB}}}\mathit{\perp }\overline{\mathit{\text{GH}}}$, $\overleftrightarrow{\mathit{\text{AB}}}\mathit{\parallel }\overleftrightarrow{\mathit{\text{CD}}}$, and $\mathit{\text{m}}\mathit{\angle }\mathit{\text{AHB=}}\mathit{60}\mathit{°}$. Find the measure of each angle.

$\angle HDE$.

Short answer

The angle $\angle HDE$ is $120°$.

Step-by-step solution

Step 1. Observe the diagram.

Here, $\angle AHD=60°$, $\overleftrightarrow{AB}\perp \overline{GH}$, $\overleftrightarrow{AB}\parallel \overleftrightarrow{CD}$ and $\overleftrightarrow{AB}$ bisects $\angle DHF$ .

Step 2. Show the calculation.

From the figure and the data,

As, $\overleftrightarrow{AB}\parallel \overleftrightarrow{CD}$, $\overleftrightarrow{AB}\parallel \overleftrightarrow{CE}$.

(As $\overleftrightarrow{CE}$is the extended part of $\overleftrightarrow{CD}$)

Take $\overleftrightarrow{HD}$as a transversal.

Use the fact that if a transversal intersects two parallel lines, then sum of each pair of co-interior angles is $180°$.

So,

$\angle AHD+\angle HDE=180°$

Put $\angle AHD=60°$.

Then,

$\begin{array}{c}{60}^{\circ }+\angle HDE={180}^{\circ }\\ \angle HDE={180}^{\circ }-{60}^{\circ }\\ ={120}^{\circ }\end{array}$

Hence the angle $\angle HDE=120°$.

Step 3. State the conclusion.

Therefore, $\angle HDE=120°$.