Question

$ABC$ is an isosceles triangle in which $AB=AC$ .$AD$ bisects exterior angle $QAC$ and $CD\parallel AB$ . Show that

(i)$\mathrm{\angle }DAC=\mathrm{\angle }BCA$
(ii)$ABCD$ is a parallelogram

Answer 1

Hint: To prove the given conditions we will use the properties of the triangles. The angles which are opposite to the equal sides must always be equal to each other. Since $AB$ and $AC$ are equal, therefore the side opposite to these sides will be equal. Also $AD$ is the bisector of the exterior angle, we will use this condition to find the relation between the angles. We will use the exterior angle property which says that the sum of the two opposite sides interior angles must always be equal to the exterior angle. In the second part of the question we need to prove that the given quadrilateral is a parallelogram for that we will prove that the pair of opposite sides in the quadrilateral $ABCD$ are parallel to each other with the help of the result of the first part.



Complete step by step solution

Given:
The triangle $\mathrm{\Delta }ABC$ is an isosceles triangle such that $AB=AC$ . Also, $AD$ bisects $\mathrm{\angle }QAC$ and $CD\parallel AB$.
(i)

Since $AD$ bisects $\mathrm{\angle }QAC$. Therefore, the angle $QAD$ and angle $DAC$ will be equal which can be expressed as:
$\mathrm{\angle }QAD=\mathrm{\angle }DAC={\displaystyle \frac{1}{2}}\mathrm{\angle }QAC$ ……(i)

Also, it is given in the question that $AB=AC$.

We know from the properties of the triangle that angles which are opposite to the equal sides must always be equal. Therefore, angle opposite to side $AB$ and $AC$ must be equal which can be expressed as:
$\mathrm{\angle }BCA=\mathrm{\angle }ABC$ ……(ii)

Now we will consider $\mathrm{\Delta }ABC$, we have angle $QAC$ is the exterior angle.

Now from the properties of the triangles, we know that the sum of the interior angles which are opposite to each other must be equal to exterior angle which can be expressed as:
$\mathrm{\angle }QAC=\mathrm{\angle }ABC+\mathrm{\angle }BCA$

We know from equation (ii) $\mathrm{\angle }BCA=\mathrm{\angle }ABC$.
$\begin{array}{l}\mathrm{\angle }QAC=\mathrm{\angle }BCA+\mathrm{\angle }BCA\\ \mathrm{\angle }QAC=2\mathrm{\angle }BCA\end{array}$

We can rewrite the above expression as:
${\displaystyle \frac{1}{2}}\mathrm{\angle }QAC=\mathrm{\angle }BCA$
We know from equation (i) $\mathrm{\angle }DAC={\displaystyle \frac{1}{2}}\mathrm{\angle }QAC$.
$\mathrm{\angle }DAC=\mathrm{\angle }BCA$

Hence, the given condition is proved.

(ii)

)

As we have proved in the first part that $\mathrm{\angle }DAC=\mathrm{\angle }BCA$ .

So, we can say that for the lines $BC$ and $AD$,$AC$ is the transversal.
But we know that $\mathrm{\angle }DAC$ and$\mathrm{\angle }BCA$ are the alternate interior angles and from the given expression they are equal. So we can conclude that lines $BC$ and $AD$ are parallel to each other which can be expressed as $BC\parallel AD$ .

Now we will consider quadrilateral $ABCD$ . As it is given in the question $CD\parallel AB$ . Also, we have proved above that $BC\parallel AD$.

Since the pair of opposite sides are parallel to each other, hence we can say that quadrilateral $ABCD$ is a parallelogram.


Note: To solve this question, we should have prior knowledge about the properties of triangles and its different angles. Here we are using the exterior angle property and angle bisector theorem. Also, we should have prior knowledge about the properties of parallelogram.