In the given figure; A E |B D, A C | E D and A B=A C. Find a , b and C.

Let P and Q be the points as shown below: Given: PDQ=58^ PDQ= EDC=58^ . .[Vertically opp .angles] EDC= ACB=58^ . . .[Corresponding angles AC | ED] In A B C, AB=AC .......[ Given ] ACB= ABC=58^ ......[angels opp. to equal sides are equal] Now, ACB+ ABC+ BAC=180^ 58^ +58^ +a=180^ a=180^ -116^ a=64^ Since A E | B D and A C is the transversal. ABC=b . . .[ Corresponding angles ] b=58^ Also since AE | BD and ED is the transversal EDC=C ....... [ Corresponding angles ] C=58^

In the given figure; A E |B D, A C | E D and A B=A C. Find a , b …

Download App

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 $]$
$\therefore \mathrm{C}=58^{\circ}$