Question

ABCD is a quadrilateral in which $AB\parallel CD$. If $AD=BC$ then show that $\angle A=\angle B$ and $\angle C=\angle D$,

Answer 1

Hint: We solve this problem by constructing a parallelogram by extending the side $AB$ to $E$ and joining with $C$. Then we use the general properties of parallelogram, parallel lines and triangle properties to prove the required results. The main properties to be used are,
(1) The angles opposite to the equal sides of a triangle are equal.
(2) The sum of interior angles between two parallel lines is ${{180}^{\circ }}$

Complete step-by-step solution:
We are given that ABCD is a quadrilateral in which $AB\parallel CD$ and $AD=BC$
Now, let us construct a quadrilateral according to the given information then we get,

Now let us construct a parallelogram by extending the side $AB$ to $E$ and then join $CE$ such that $AD\parallel CE$ then we get,

Now, let us assume the parallel lines $AE\parallel DC$ and $CE$ is the transversal.
We know that the sum of two interior angles of parallel lines equal to ${{180}^{\circ }}$
By using this condition to $\angle A$ and $\angle E$ then we get,
$\begin{align}
  & \Rightarrow \angle A+\angle E={{180}^{\circ }} \\
 & \Rightarrow \angle A={{180}^{\circ }}-\angle E...........equation(i) \\
\end{align}$
We know that the quadrilateral AECD is a parallelogram and in a parallelogram the opposite sides are equal.
By using this property of quadrilateral ACD then we get,
$\Rightarrow AD=CE$
We are given that $AD=BC$ so that the above equation can be written as,
$\Rightarrow BC=CE$
Now, let us consider the triangle $\Delta BCE$
We know that the property of triangle that is the angles opposite to equal sides are equal.
By using this condition to triangle $\Delta BCE$ we get,
$\begin{align}
  & \Rightarrow BC=CE \\
 & \Rightarrow \angle CBE=\angle CEB \\
\end{align}$
Now, let us consider the vertex $B$ at which there are two angles in a straight line.
We know that the sum of angles cut in a straight is equal to ${{180}^{\circ }}$
By using this condition at the vertex $B$ then we get,
\[\begin{align}
  & \Rightarrow \angle CBA+\angle CBE={{180}^{\circ }} \\
 & \Rightarrow \angle CBA+\angle CEB={{180}^{\circ }} \\
\end{align}\]
Now, let us represent the angle $\angle CBA=\angle B$ and $\angle CEB=\angle E$ then we get,
$\Rightarrow \angle B={{180}^{\circ }}-\angle E$
Now, by replacing the required value from equation (i) in above equation then we get,
$\Rightarrow \angle B=\angle A...........equation(ii)$
We know that in a parallelogram the opposite angles are equal.
By using this condition to parallelogram $AECD$ then we get,
$\Rightarrow \angle A=\angle C........equation(iii)$
Now, let us consider the diagonal $DB$.

Here, we can see that $AB\parallel DC$ and $DB$ is the transversal.
We know that the corresponding angles in parallel lines are equal. By using this condition we get,
$\begin{align}
  & \Rightarrow \angle BDC=\angle DBA \\
 & \Rightarrow \angle BDA=\angle DBC \\
\end{align}$
Now let us add the above two equations then we get,
$\begin{align}
  & \Rightarrow \angle BDC+\angle BDA=\angle DBA+\angle DBC \\
 & \Rightarrow \angle D=\angle B..........equation(iv) \\
\end{align}$
Now, let us substitute the required values in equation (ii) from the equation (iii) and (iv) then we get,
$\Rightarrow \angle C=\angle D$
Therefore, we can conclude that the required result has been proved that is,
$\begin{align}
  & \therefore \angle A=\angle C \\
 & \therefore \angle C=\angle D \\
\end{align}$

Note: We need to be careful in taking the required properties in order to prove the required result. Here, we can see that we used many standard properties to prove the required answer. One can make mistakes in taking the standard property in the wrong manner or else one can forget the required property that leads to the correct answer. So, we need to remember the standard properties that are very useful.
Here, we can see that the properties used are,
(1) In a triangle the opposite angles of equal sides are equal.
(2) The sum of interior angles between two parallel lines and the transversal is ${{180}^{\circ }}$ and the corresponding angles are equal.
(3) Sum of angles cut in a straight line is equal to ${{180}^{\circ }}$