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