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

Question 8: ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD. Show that

- `ΔAPB ≅ ΔCQD`
- `AP = CQ`

**Answer:** In ΔAPB and ΔCQD

`∠ABP=∠CDQ` (alternate angles of transversal DB)

`AB=CD`

`∠APB=∠CQD` (right angles)

Hence, `ΔAPB≅ΔCQD`

So, `AP=CQ`

Question 9: In ∆ ABC and ∆ DEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively. Show that

- quadrilateral ABED is a parallelogram
- quadrilateral BEFC is a parallelogram
- AD || CF and AD = CF
- quadrilateral ACFD is a parallelogram
- AC = DF
- ΔABC ≅ ΔDEF

**Answer:** In ΔABC and ΔDEF

`AB=DE` (given)

`BC=EF` (given)

`∠ABC=∠DEF` (AB||DE and BC||EF)

Hence, `ΔABC≅ΔDEF`

In quadrilateral ABED

AB= ED

AB||ED

So, ABED is a parallelogram (opposite sides are equal and parallel)

So, BE||AD ------------ (1)

Similarly quadrilateral ACFD can be proven to be a parallelogram

So, BE||CF ------------ (2)

From equations (1) & (2)

It is proved that

AD||CF

So, AD=CF

Similarly AC=DF and AC||DF can be proved

Question 10: ABCD is a trapezium in which AB || CD and AD = BC. Show that

- `∠A=∠B`
- `∠C=∠D`
- `ΔABC ≅ ΔBAD`
- Diagonal AC = Diagonal BD

**Answer:** In ΔBCE
`EC=AD` (opposite sides are equal in parallelogram)

`AD=BC` (given)

So, `BC=EC`

So, `∠CBE=∠CEB`

`∠CBE+∠CBA=180°` (linear pair of angles)

`∠CEB+∠DAB=180°` (adjacent angles of parallelogram are complementary)

substituting `∠CBE=∠CEB` it is clear that `∠DBA=∠CBA`

Now, `∠DAB+∠CDA=180°` (adjacent angles of parallelogram)

And, `∠CBA+∠DCB=180°` (adjacent angles of parallelogram)

As `∠DBA=∠CBA` so it is clear that `∠CDA=∠DCB`

In ΔABC and ΔBAD

`AB=AB` (common side)

`AD=BC` (given)

`∠DBA=∠CBA`

hence, `ΔABC≅ΔBAD`

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