Question

In parallelogram $ ABCD$, two points $ P$ and $ Q$ are taken on diagonal $ BD$ such that $ DP = BQ$. Show that:

Answer 1

Step 1: Show that $△APD$ and $△CQB$ are congruent

It is given that $BD$ is the diagonal of the parallelogram $ABCD$.

Also, given that $DP=BQ$.

Now, in $△APD$ and $△CQB$,

$AD=CB$ [opposite sides of the parallelogram $ABCD$]

$\angle PDA=\angle QBC$ [alternate interior angles]

$PD=QB$ [given]

Then by SAS (Side-Angle-Side) congruency axiom,

$△APD\cong △CQB$

Hence, it is proved that $△APD\cong △CQB$.

Step 2: Use the CPCTC rule to equate $AP$ and $CQ$

As we know, the corresponding parts of the congruent triangles are congruent (CPCTC). So,

$\because △APD\cong △CQB$

$\Rightarrow$ $AP=CQ$ …(i)

Hence, it is proved that $AP=CQ$

Step 3: Show that $△AQB$ and $△CPD$ are congruent

In $△AQB$ and $△CPD$,

$BA=DC$ [opposite sides of the parallelogram $ABCD$]

$\angle QBA=\angle PDC$ [alternate interior angles]

$QB=PD$ [given]

Then by SAS (Side-Angle-Side) congruency axiom,

$△AQB\cong △CPD$

Hence, it is proved that $△AQB\cong △CPD$.

Step 4: Use the CPCTC rule to equate $AQ$ and $CP$

As we know, the corresponding parts of the congruent triangles are congruent (CPCTC). So,

$\because △AQB\cong △CPD$

$\Rightarrow$ $AQ=CP$ …(ii)

Hence, it is proved that$AQ=CP$.

Step 5: Show that $APCQ$ is a parallelogram

From equations (i) and (ii), we have

$AP=CQ$ and $AQ=CP$

In a quadrilateral$APCQ$, the opposite sides are equal.

So, opposite angles will also be equal.

$\Rightarrow APCQ$ is a parallelogram.

Hence, it is proved that $APCQ$ is a parallelogram.

Hence, it is proven that

1. $△APD\cong △CQB$
2. $AP=CQ$
3. $△AQB\cong △CPD$
4. $AQ=CP$
5. $APCQ$ is a parallelogram.