Class 9 Maths

Quadrilaterals

Exercise 8.1 Part 3

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`
quadrilateral

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
quadrilateral

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
quadrilateral

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`