# Quadrilaterals

## Exercise 8.1 Part 2

Question 5: Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

**Answer:** Using the same figure,

If DO=AO

Then `∠DAO=∠BAO=45°`

(Angles opposite to equal sides are equal)

So, all angles of the quadrilateral are right angles making it a square.

Question 6: Diagonal AC of a parallelogram ABCD bisects angle A . Show that

(i) it bisects angle C also,

(ii) ABCD is a rhombus.

**Answer:** ABCD is a parallelogram where diagonal AC bisects angle DAB

In ΔADC and ΔABC

`∠DAC=∠`BAC (diagonal is bisecting the angle)

`AC=AC` (common side)

`AD=BC` (parallel sides are equal in a parallelogram)

Hence, `ΔADC≅ΔABC`

So, `∠DCA=∠BCA`

This proves that AC bisects ∠DCB as well.

Now let us assume another diagonal DB intersecting AC on O.

As it is a parallelogram so DB will bisect AC and vice-versa.

In ΔAOD and ΔBOD

`∠DAO=∠DCO`

(opposite angles are equal in parallelogram so their halves will be equal)

`AO=CO`

`DO=DO`

Hence `ΔAOD ≅ ΔBOD`

So, `∠DOA=∠DOB=90°`

As diagonals are intersecting at right angles so it is a rhombus

Question 7: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ. Show that:

- `ΔAPD≅ ΔCQB`
- `AP=CQ`
- `ΔAQB≅ΔCPD`
- `AQ=CP`
- APCQ is a parallelogram

**Answer:** In ΔAPD and ΔCQB

`DP=BQ` (given)

`AD=BC` (opposite sides are equal)

`∠DAP=∠BCQ` (opposite angles’ halves are equal)

Hence, `ΔAPD≅ΔCQB`

So, `AP=CQ` proved

In ΔAQB and ΔCPD

`AB=CD` (opposite sides are equal)

`DP=BQ` (given)

`∠BAQ=∠DCP` (opposite angles’ halves are equal)

Hence, `ΔAQB≅ΔCPD`

So, `AQ=CP` proved

`∠DPA=∠BQP`

(corresponding angles of congruent triangles APD and CQB)

In ΔDQP and ΔBQP

`∠DPQ=∠BQP`

(from previous proof)

`DP=BQ` (given)

`PQ=PQ` (common side)

So, `ΔDQP≅ΔBQP`

So, `∠QDP=∠QBP`

With equal opposite angles and equal opposite sides it is proved that APCQ is a parallelogram