Question

In a parallelogram ABCD, two points P and Q are taken on diagonal BD such that $DP=BQ$(see the given figure). Show that:

(i) $\Delta APD\cong \Delta CQB$ \[\]
(ii) $AP=CQ$ \[\]
(iii) $\Delta AQB\cong \Delta CPD$ \[\]
(iv) $AQ=CP$ \[\]
(v) APCQ is a parallelogram \[\]

Answer 1

Hint: We take equal alternate angles $\angle ADP=\angle CBQ$, opposite sides $AD=BC$ and the given equality $DP=BQ$ to prove $\Delta APD\cong \Delta CQB$ which gives us $AP=CQ$. Similarly we take equal alternate angles $\angle ABQ=\angle CDP$, opposite sides $AB=BD$ the given equality $DP=BQ$ to prove $\Delta AQB\cong \Delta CPD$ which gives us $AQ=CP$. The proofs of part (ii) and (iv) gives that APCQ is a parallelogram \[\]

Complete step-by-step solution
We know that the opposite sides of a parallelogram are equal and parallel. In parallelogram ABCD we have opposite sides AB and CD , AD and BC. Then we have;
\[\begin{align}
  & AB=CD,AB||CD \\
 & AD=BC,AD||BC \\
\end{align}\]
We also know that the measure of opposite angles in a parallelogram is equal. \[\]
(i) Let us observe the triangles APD and CBQ. Since we have $AD||BC$ and the diagonal BD as the transverse line, we have equal alternate angles$\angle ADP=\angle CBQ$. The opposite sides of parallelogram $AD=BC$ and we are given equal sides $DP=BQ$. We use side-angle-side congruence and conclude that $\Delta APD\cong CQB$. \[\]

(ii) Since we have proved $\Delta APD\cong CQB$ the corresponding sides will be equal and hence sides opposite to $\angle ADP=\angle CBQ$ will be equal which means $AP=CQ$ \[\]

(iii) Let us observe the triangles AQB and CPD. We have the parallel sides of parallelogram $AB||CD$ and the diagonal BD as the transversal line where we have equal alternating angles$\angle ABQ=\angle CDP$. We have equal opposite sides $AB=CD$ and we are given in the question $DP=BQ$. We use side-angle-side congruence and conclude that $\Delta AQB\cong \Delta CPD$. \[\]

(iv) Since we have proved $\Delta AQB\cong \Delta CPD$ the corresponding sides will be equal and hence sides opposite to $\angle ABQ=\angle CDP$ will be equal which means $AQ=CP$ \[\]
(v)Since we have already proved $AP=CQ$ and $AQ=CP$ which are opposite sides of the quadrilateral APCQ, then APCQ is a parallelogram. \[\]

Note: We note that both pair of opposite sides have to be equal or parallel to make the quadrilateral a parallelogram; otherwise if one pair of sides are equal or parallel the quadrilateral is a trapezium. We can also directly prove APCQ is a parallelogram. We must not confuse between the similarity and congruence of a triangle if two triangles are similar or may not be congruent.