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.

quadrilaterals 8

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:

  1. `ΔAPD≅ ΔCQB`
  2. `AP=CQ`
  3. `ΔAQB≅ΔCPD`
  4. `AQ=CP`
  5. APCQ is a parallelogram
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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

  1. `ΔAPB ≅ ΔCQD`
  2. `AP = CQ`
quadrilaterals 14

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

  1. quadrilateral ABED is a parallelogram
  2. quadrilateral BEFC is a parallelogram
  3. AD || CF and AD = CF
  4. quadrilateral ACFD is a parallelogram
  5. AC = DF
  6. ΔABC ≅ ΔDEF
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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

  1. `∠A=∠B`
  2. `∠C=∠D`
  3. `ΔABC ≅ ΔBAD`
  4. Diagonal AC = Diagonal BD
quadrilaterals 20

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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