Question
In the given figure. $DE \| BC$. If $\angle\text{C}=65^\circ$ and $\angle\text{B}=55^\circ$, find
$i. \angle\text{ADE}$
$ii. \angle\text{AED}$
$iii. \angle\text{C}$
$i. \angle\text{ADE}$
$ii. \angle\text{AED}$
$iii. \angle\text{C}$
✓
Answer
$i. DE \| BC$
$\therefore\angle\text{ABC}=\angle\text{ADE}=55^\circ ($corresponding angles$)$
$ii.$ Sum of the angles of any triangle is $180^\circ$.
$\therefore\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\angle\text{C}=180^\circ-(65^\circ-55^\circ)=60^\circ$
$DE \| BC$
$\therefore \angle\text{AED}=\angle\text{ACB}=60^\circ ($corresponding angles$)$
$iii.$ We have found in point $(ii)$ that $\angle\text{C}=60^\circ$
$\therefore\angle\text{ABC}=\angle\text{ADE}=55^\circ ($corresponding angles$)$
$ii.$ Sum of the angles of any triangle is $180^\circ$.
$\therefore\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\angle\text{C}=180^\circ-(65^\circ-55^\circ)=60^\circ$
$DE \| BC$
$\therefore \angle\text{AED}=\angle\text{ACB}=60^\circ ($corresponding angles$)$
$iii.$ We have found in point $(ii)$ that $\angle\text{C}=60^\circ$
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