Class 9 Maths

Congurency In Triangles

Exercise 7.1

Question 1: In quadrilateral ABCD, `AC = AD` and AB bisects ∠A. Show that ΔACB ≅ ΔABD.

Quadrilateral

Answer: in ΔACB and ΔABD
`AC=AD`
`∠CAB=∠DAB` (AB is bisecting ∠CAD)
`AB=AB` (common side)
So, SAS axiom it is proved that:
ΔACB ≅ ΔABD

Question 2: ABCD is a quadrilateral in which `AD=BC` and `∠DAB=∠CAB`

Prove that

  1. `ΔABD≅ ΔBAC`
  2. `BD=AC`
  3. `∠ABD=∠BAC`
Quadrilateral

Answer: In ΔABD and ΔBAC
AD=BC
AB=AB (common side)
∠BAD=∠ABC
So, by SAS rule ΔABD∝ ΔBAC

Since ΔABD≅ ΔBAC
So, BD=AC
(Third corresponding sides of respective tri∠s)

In congruent triangles all corresponding angles are always equal.
So, ∠BAD=∠ABC proved

Question 3: AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB.

Triangle

Answer: In ΔBOC and ΔAOD
`BC=AD` (given)
`∠CBO=∠DAO` (right angle)
`∠BOC=∠AOD` (opposite angles)
So, by ASA rule
`ΔBOC≅ ΔAOD`
Or, `BO=AO`
And it is proved that CD bisects AB.

Question 4: l and m are two parallel lines intersected by another pair of parallel lines p and q. Show that

ΔABC ≅ Δ CDA

Quadrilateral

Answer: In ΔABC and ΔCDA
`AB=CD` (l and m are parallel)
`AD=BC` (AB and CD are parallel)
`∠ABC=∠DCm` (angles on the same side of transversal BC)
`∠DCm=∠ADC`
So, by SAS rule `ΔABC≅ΔCDA`

Question 5: Line l is the bisector of an angle A and B is any point on l. BP and BQ are perpendiculars from B to the arms of angle A. Show that:

  1. ΔAPB ≅ ΔAQB
  2. BP = BQ or B is equidistant from the arms of  angle A.
Lines

Answer: In ΔAPB and ΔAQB
`AB=AB` (common side)
`∠PAB=∠QAB` (AB is bisector of ∠QAP)
`∠AQB=∠APB` (right angle)
So, by ASA rule ΔAPB ≅ ΔAQB
And `BQ=BP`

Question 6: In the given figure, AC = AE, AB = AD and angle BAD = angle EAC. Show that BC = DE.

Quadrilateral

Answer: In ΔABC and ΔADE
`AB=AD` (given)
`AC=AE` (given)
Since `∠BAD=∠EAC`
So, `∠BAD+∠DAC=∠EAC+∠DAC`
Or, `∠BAC=∠DAE`
So, by SAS rule `ΔABC ≅ ΔADE`
Or, `BC=DE` proved

Question 7: AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that angle BAD = angle ABE and angle EPA = angle DPB Show that

  1. `ΔDAP≅ ΔEBP`
  2. `AD=BE`
Triangle

Answer: In ΔDAP and ΔEBP
`∠BAD=∠ABE` (given)
`∠EPA=∠DPB` (given)
So, `∠EPA+∠EPD=∠DPB+∠EPD`
Or, `∠DPA=∠EPB`
`AP=PB` (P is midpoint of AB)
So, by ASA rule `ΔDAP≅ ΔEBP`
So, `AD=BE`

Question 8: In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B. Show that:

  1. `ΔAMC ≅ ΔBMD`
  2. `∠DBC` is a right angle
  3. `ΔDBC ≅ ΔACB`
  4. `CM=1/2\AB`
Triangles

Answer: In ΔAMC and ΔBMD
`BM=AM` (M is midpoint)
`DM=CM` (given)
`∠DMB=∠AMC` (opposite angles)
So, `ΔAMC≅ ΔBMD`
Hence, `DB=AC`
`∠DBA=∠BAC`
So, DB||AC (alternate angless are equal)
So, `∠BDC=∠ACB=` Right angle
(Internal angles are complementary in case of transverse of parallel lines)

In ΔDBC and ΔACB
`DB=AC` (proven earlier)
`BC=BC` (common side)
∠BDC=∠ACB (proven earlier)
So, `ΔDBC≅ ΔACB`
So, `AB=DC`
So, `AM=BM=CM=DM`
So, `CM=1/2\AB`


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