Question

In an isosceles triangle ABC, $AC=BC$, $\angle BAC$ is bisected by AD where D lies on BC. It is found that $AD=AB$. Then $\angle ACB$ equals

A) $72{}^\circ $
B) $54{}^\circ $
C) $36{}^\circ $
D) $60{}^\circ $

Answer 1

Hint: First assume the value of $\angle BAC$. As $AC=BC$, so the opposite angles are equal. So, the assumed value will be equal to \[\angle ABC\]. Now in triangle BAD, $AD=AB$, again opposite angles will be equal. So, the assumed angle will also be equal to \[\angle ADB\]. As AD is bisector, then \[\angle BAD\] will be half of the assumed angles. Now find the assumed angle value by adding all angles of triangle BAD which will be equal to $180{}^\circ $. After that in triangle ABC, substitute the assumed angle value and find \[\angle ACB\].

Complete step-by-step answer:
Given: - $AB=BC$.
AD is the bisector of $\angle BAC$.
$AD=AB$
Let $\angle BAC$ be x.
In triangle ABC, $AB=AC$. Then,
$\angle BAC=\angle ABC$ (opposite angles equal)
Substitute $\angle BAC=x$,
$\angle ABC=x$ ….. (1)
In triangle BAD, $AD=AB$. Then,
$\angle ADB=\angle DBA$ (opposite angles equal)
Since $\angle DBA$ and $\angle ABC$ is the same angle. Substitute the value of $\angle DBA$ from equation (1),
$\angle ADB=x$
Since AD is the bisector of $\angle BAC$. So,
$\angle BAD=\dfrac{x}{2}$
As we know that the sum of angles of a triangle is equal to $180{}^\circ $. Then,
$\angle BAD+\angle BDA+\angle DBA$
Substitute the values of angles,
$\dfrac{x}{2}+x+x=180{}^\circ $
Take LCM on the left side,
$\dfrac{x+2x+2x}{2}=180{}^\circ $
Add the terms in the numerator and multiply the denominator on the right side,
$5x=360{}^\circ $
Divide both sides by 5,
$x=72{}^\circ $
Now, in triangle ABC, the sum of angles is equal to $180{}^\circ $,
$\angle ABC+\angle BAC+\angle ACB=180{}^\circ $
Substitute the value of $\angle ABC$ and $\angle BAC$ in the equation,
$72{}^\circ +72{}^\circ +\angle ACB=180{}^\circ $
Add the angles on the left side,
$144{}^\circ +\angle ACB=180{}^\circ $
Move the angle value on the right side and subtract it from $180{}^\circ $,
$\angle ACB=36{}^\circ $
Thus, the value of $\angle ACB$ is $36{}^\circ $.

Hence, option (C) is correct.

Note: A “triangle” is a three-sided polygon having three angles. The sum of all interior angles of a triangle will always add up to 180 degrees. This is called the angle sum property of the triangle.
The sum of all the internal angles of a triangle is always 180o no matter how the triangle is constructed.
The length of any side of a triangle is shorter than the sum of the other two sides.
A triangle can always be split into two right triangles no matter how the triangle is constructed.