Question
In the given figure; $A E\|B D, A C\| E D$ and $A B=A C$. Find $\angle a , \angle b$ and $\angle C$.
✓
Answer
Let $\mathrm{P}$ and $\mathrm{Q}$ be the points as shown below:
Given:
$\angle \mathrm{PDQ}=58^{\circ}$
$ \angle \mathrm{PDQ}=\angle \mathrm{EDC}=58^{\circ} \ldots . .[$Vertically opp .angles$]$
$ \angle \mathrm{EDC}=\angle \mathrm{ACB}=58^{\circ} \ldots . . .[$Corresponding angles $\because \mathrm{AC} \| \mathrm{ED}]$
In $\triangle A B C$,
$\mathrm{AB}=\mathrm{AC}\dots.......[$ Given $]$
$\therefore \angle \mathrm{ACB}=\angle \mathrm{ABC}=58^{\circ}\dots......[$angels opp. to equal sides are equal$]$
Now,
$\angle \mathrm{ACB}+\angle \mathrm{ABC}+\angle \mathrm{BAC}=180^{\circ}$
$ \Rightarrow 58^{\circ}+58^{\circ}+\mathrm{a}=180^{\circ}$
$ \Rightarrow \mathrm{a}=180^{\circ}-116^{\circ}$
$ \Rightarrow \mathrm{a}=64^{\circ}$
Since $A E \| B D$ and $A C$ is the transversal.
$\angle \mathrm{ABC}=\mathrm{b} \ldots . . .[$ Corresponding angles $]$
$\therefore \mathrm{b}=58^{\circ}$
Also since $\mathrm{AE}\| \mathrm{BD}$ and $\mathrm{ED}$ is the transversal $\angle \mathrm{EDC}=\mathrm{C}\dots....... [$ Corresponding angles $]$
$\therefore \mathrm{C}=58^{\circ}$
Given:
$\angle \mathrm{PDQ}=58^{\circ}$
$ \angle \mathrm{PDQ}=\angle \mathrm{EDC}=58^{\circ} \ldots . .[$Vertically opp .angles$]$
$ \angle \mathrm{EDC}=\angle \mathrm{ACB}=58^{\circ} \ldots . . .[$Corresponding angles $\because \mathrm{AC} \| \mathrm{ED}]$
In $\triangle A B C$,
$\mathrm{AB}=\mathrm{AC}\dots.......[$ Given $]$
$\therefore \angle \mathrm{ACB}=\angle \mathrm{ABC}=58^{\circ}\dots......[$angels opp. to equal sides are equal$]$
Now,
$\angle \mathrm{ACB}+\angle \mathrm{ABC}+\angle \mathrm{BAC}=180^{\circ}$
$ \Rightarrow 58^{\circ}+58^{\circ}+\mathrm{a}=180^{\circ}$
$ \Rightarrow \mathrm{a}=180^{\circ}-116^{\circ}$
$ \Rightarrow \mathrm{a}=64^{\circ}$
Since $A E \| B D$ and $A C$ is the transversal.
$\angle \mathrm{ABC}=\mathrm{b} \ldots . . .[$ Corresponding angles $]$
$\therefore \mathrm{b}=58^{\circ}$
Also since $\mathrm{AE}\| \mathrm{BD}$ and $\mathrm{ED}$ is the transversal $\angle \mathrm{EDC}=\mathrm{C}\dots....... [$ Corresponding angles $]$
$\therefore \mathrm{C}=58^{\circ}$
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