Question

In the given figure, lines (i) and (ii) are parallel lines and $AC$ is the angle bisector of $\angle BAD$.

What type of triangle is $△ABC$?

• A. Equilateral Triangle
• B. Acute Angled Triangle
• C. Isosceles Triangle
• D. Scalene Triangle

Answer 1

Answer: (A), (B), (C)

The correct options are
A

Equilateral Triangle

B

Acute Angled Triangle

C

Isosceles Triangle

In the figure,

$\because \angle CBE={120}^{\circ }$

$\therefore \angle BAD={120}^{\circ }$ [Corresponding angles]

Since, $\angle BAD={120}^{\circ }$

$\Rightarrow \angle BAC=\angle CAD=\frac{\angle BAD}{2}=\frac{120}{2}$ = ${60}^{\circ }$ [Because $AC$ is the angle bisector of $\angle BAD$]

Also,

$\angle ABC={180}^{\circ }$ - $\angle CBE$ = ${180}^{\circ }$ - ${120}^{\circ }$ = ${60}^{\circ }$ [Linear pair]

In $△ABC,\angle A+\angle B+\angle C={180}^{\circ }$ [Angle sum property]

$\Rightarrow \angle C={180}^{\circ }$- $\angle A-\angle B$

$=$ ${180}^{\circ }$ - ${60}^{\circ }$ - ${60}^{\circ }$ = ${60}^{\circ }$

$\Rightarrow \angle C={60}^{\circ }$

Thus, all the angles of $△ABC={60}^{\circ }$

Hence, $△ABC$ is an equilateral triangle which is also acute angled.

Since all the equilateral triangles are isosceles, $△ABC$ is an isosceles triangle as well.

So, options A, B and C are correct