Question

In parallelogram $ABCD$ two point $P$ and $Q$ are taken on diagonal $BD$ such that $DP = BQ$ . Show that
$
  \left( {\text{i}} \right){\text{ }}\vartriangle APD \cong \vartriangle CQB \\
  \left( {{\text{ii}}} \right){\text{ }}AP = CQ \\
  \left( {{\text{iii}}} \right){\text{ }}\vartriangle AQB \cong \vartriangle CPD \\
  \left( {{\text{iv}}} \right){\text{ }}AQ = CP \\
  \left( {\text{v}} \right){\text{ }}APCQ{\text{ is a parallelogram }} \\
 $

Answer 1

Hint – In this question, firstly we have to analyse the question and we have to write what is given to us i.e. ${\text{ABCD}}$ is a parallelogram with a diagonal draw between ${\text{B and D}}$ and on this diagonal we also take two points ${\text{P and Q}}$ where ${\text{DP = BQ}}$. Now, this will give us a clear picture to understand the question. Eventually, we will consider all the given options and prove them one by one. So, that we can get the desired result. As in this question, we have to prove all the given points one after another.

Complete step-by-step solution -
Given - ${\text{ABCD}}$ is a parallelogram with diagonal ${\text{BD}}$ with points ${\text{P and Q}}$ on them where ${\text{DP = BQ}}$.
To prove - $\vartriangle {\text{APD}} \cong \vartriangle {\text{CQB}}$
Proof – Now, $\left. {{\text{AD}}} \right\|{\text{BC}}$ (As you know that , opposite sides of parallelogram are parallel )
and ${\text{BD}}$ is transversal.
$\angle {\text{ADP = }}\angle {\text{CBQ}}$ (As they are alternate angles alongside diagonal ${\text{BD}}$) – (1)
Now, in$\vartriangle {\text{APD and }}\vartriangle {\text{CQB}}$.
${\text{AD = CB}}$ (Now, we know that opposite sides of parallelogram are equal)
$\angle {\text{ADP = }}\angle {\text{CBQ}}$ (Alternate angles, proven in (1))
 ${\text{DP = BQ}}$ (It is given in question)
So, now we can say that $\vartriangle {\text{APD}} \cong \vartriangle {\text{CQB}}$ i.e. $\vartriangle {\text{APD and }}\vartriangle {\text{CQB}}$ are congruent triangles by SAS congruence rule.
To prove - $\left( {{\text{ii}}} \right){\text{ AP = CQ}}$
Proof - Now, in previous part we proved that $\vartriangle {\text{APD}} \cong \vartriangle {\text{CQB}}$, therefore we can say that ${\text{AP = CQ}}$. As corresponding parts of congruent triangles are equal.
To prove - $\left( {{\text{iii}}} \right)$ $\vartriangle {\text{AQB}} \cong \vartriangle {\text{CPD}}$
Proof – Since, $\left. {{\text{AB}}} \right\|{\text{DC}}$ (As you know that , opposite sides of parallelogram are parallel )
and ${\text{BD}}$ is transversal.
$\angle {\text{ABQ = }}\angle {\text{CDP}}$ (As they are alternate angles alongside diagonal ${\text{BD}}$) – (1)
Now, in $\vartriangle {\text{AQB and }}\vartriangle {\text{CPD}}$.
${\text{AB = CD}}$ (Now, we know that opposite sides of parallelogram are equal)
$\angle {\text{ABQ = }}\angle {\text{CDP}}$ (Alternate angles, proven in (1))
${\text{BQ = DP}}$ (It is given in question)
So, now we can say that $\vartriangle {\text{AQB}} \cong \vartriangle {\text{CPD}}$ i.e. $\vartriangle {\text{AQB and }}\vartriangle {\text{CPD}}$ are congruent triangles by SAS congruence rule.
To prove - $\left( {{\text{iv}}} \right)$ ${\text{AQ = CP}}$
Proof - Now, in previous part we proved that $\vartriangle {\text{AQB}} \cong \vartriangle {\text{CPD}}$ therefore we can say that ${\text{AQ = CP}}$. As corresponding parts of congruent triangles are equal.
To prove - $\left( {\text{v}} \right)$ ${\text{APCQ}}$ is a parallelogram
Proof – As, we know that in part $\left( {{\text{ii}}} \right){\text{ and }}\left( {{\text{iv}}} \right)$ we proved that ${\text{AP = CQ and AQ = CP}}$.
Since, both pairs of opposite sides in ${\text{APCQ}}$ are equal. So, we can say that ${\text{APCQ}}$ is a parallelogram.
Hence, we proved all the parts.

Note – Eventually, we can say that we learnt about parallelogram.
Parallelogram: A quadrilateral in which both pairs of opposite sides are parallel is called a parallelogram
A quadrilateral is a parallelogram if:
$\left( {\text{i}} \right)$ Its opposite sides are equal.
$\left( {{\text{ii}}} \right)$ Its opposite angles are equal.
$\left( {{\text{iii}}} \right)$ Diagonals bisect each other.
$\left( {{\text{iv}}} \right)$ A pair of opposite sides is equal and parallel.
Henceforth, whenever we face such types of problems you should lay emphasis on what is given and then from that given part you should prove the parts one by one respectively because there must be a probability that all the parts are connected. So, it is recommended you should solve them in a particular given order. So, use this approach as your key concept like we did in this question. Thus, this approach will lead us to our required answer.