BIGtheme.net http://bigtheme.net/ecommerce/opencart OpenCart Templates
Tuesday , August 22 2017
Home / Geometry / Exercise 10.1|Class seven| Congruence| Part – 1

# Exercise 10.1|Class seven| Congruence| Part – 1

Problem – 1: In the figure, CD is the perpendicular bisector of AB, Prove that ∆ADC ≅ ∆BDC.

Solution: Particular enunciation: Given that, in the figure, CD is the perpendicular bisector of AB. i.e., AD = BD. We have to prove that, ∆ADC ≅ ∆BDC.

CD is the common side and ∠ADC = ∠BDC = 900 [∵ CD ⊥ AB]

Problem – 2: In the figure, CD = CB and ∠DCA = ∠BCA. Prove that AB = AD.

Solution: Particular enunciation: Given that, in the figure, CD = CB and ∠DCA = ∠BCA. We have to prove that AB = AD.

Proof: In ∆ABC and ∆ADC, we get

CB = CD

AC is the common side of both triangles,

and ∠DCA = ∠BCA

Problem – 3: In the figure, ∠BAC = ∠ACD and AB = DC. Prove that AD = BC, ∠CAD = ∠ACD and ∠ADC = ∠ABC.

Solution: Given that, in the figure, ∠BAC = ∠ACD and AB = DC. We have to prove that AD = BC, ∠CAD = ∠ACD and ∠ADC = ∠ABC.

AB = DC

AC is the common side

and included ∠BAC = ∠ACD

Problem – 4: If the base of an isosceles triangle is produced both ways, show that the exterior angles so formed are equal.

Solution: General enunciation: If the base of an isosceles triangle is produced both ways, show that the exterior angles so formed are equal.

Particular enunciation: Let, ∆ABC is an isosceles triangle where AB = AC and BC its base. The base BC is produced on both sides up to P and Q and exterior angles so formed are ∠ABQ and ∠ACP.

Proof: In ∆ABC

AB = AC

∴∠ACB = ∠ABC

Since ∠ABC and ∠ABQ are adjacent angles and lie on the same line,

∴ ∠ABC + ∠ABQ = 1800 —————————- (1)

And ∠ACB + ∠ACP = 1800 ————————- (2)

From equation (1) and (2) we get,

∠ABC + ∠ABQ = ∠ACB + ∠ACP

⟹ ∠ABC + ∠ABQ = ∠ABC + ∠ACP

⟹ ∠ABQ = ∠ACP [neglect ∠ABC from both side]

∴ ∠ABQ = ∠ACP.    (Proved)

## If two angles of triangles are equal, then the sides opposite to the equal angles are equal.

If two angles of triangles are equal, then the sides opposite to the equal angles ...