Question

ABCD is a parallelogram and AP and CQ are perpendicular from vertices A and C on diagonal BD. Show that,
$\left( i \right)\Delta APB\cong \Delta CQD$
(ii) AP = CQ

Answer 1

Hint: In this question, we are given a figure in which ABCD is a parallelogram. We need to prove congruence of two triangles $\Delta APB\text{ and }\Delta CQD$. Also we need to prove AP = CQ. For this, we will use the angle angle side (AAS) congruence rule. We will use the property that alternate interior angles are equal and the opposite sides of a parallelogram are equal. After that, we will use the property of CPCT (corresponding parts of congruent triangle) to prove AP to be equal to CQ.

Complete step by step answer:
Here we are given the diagram as,

ABCD is a parallelogram and BD is one of its diagonal. AP and CQ are two perpendicular drawn from vertices A and C on the diagonal BD.
Now we know that, in a parallelogram, opposite sides are equal. So in this parallelogram, we can say that, AB is equal to CD i.e. $AB=CD\ldots \ldots \ldots \left( 1 \right)$.
Also, in a parallelogram, opposite sides are parallel, therefore we can say that AB is parallel to CD.
We can see from the diagram that, $AB\parallel CD$ with BD acting as in transversal. Now, $\angle CDB\text{ and }\angle ABD$ are the angles made by parallel lines with transversal which are alternate interior and we know that alternate interior angles are equal. Therefore, for $AB\parallel CD$ and BD as transversal,
$\angle CDB=\angle ABD\left( \text{alternate interior angles} \right)\ldots \ldots \ldots \left( 2 \right)$.
Now, we are given that, AP and CQ are perpendicular on the diagonal BD. Therefore, they make an angle of ${{90}^{\circ }}$ with BD. Since both $\angle APB\text{ and }\angle CQD$ are ${{90}^{\circ }}$ so we can say that $\angle APB=\angle CQD\ldots \ldots \ldots \left( 3 \right)$.
Let us look at the two triangles $\Delta APB\text{ and }\Delta CQD$.
As found earlier from (1), (2) and (3) we find that,
$\begin{align}
  & AB=CD \\
 & \angle CDB=\angle ABD \\
 & \angle APB=\angle CQD \\
\end{align}$
Therefore, by AAS congruence criteria, we can say that $\Delta APB\cong \Delta CQD$.
Now we know that, corresponding parts of two congruent triangles (CPCT) are equal, so from $\Delta APB\text{ and }\Delta CQD$ we can say that AP = CQ.

Note: Students should keep in mind all the congruence rules that can be applied before solving this sum. They should note that AAA (Angle angle angle) is not a congruence condition. By CPCT we can also prove the other corresponding sides and angles are equal.