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.

Proof: In ∆ADC and ∆BDC, AD = BD

CD is the common side and ∠ADC = ∠BDC = 90^{0} [∵ CD ⊥ AB]

∴ ∆ADC ≅ ∆BDC (Proved)

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

∴ ∆ABC ≌ ∆ADC

∴ AB = AD (Proved)

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.

Proof: In ∆ABC and ∆ADC,

AB = DC

AC is the common side

and included ∠BAC = ∠ACD

∴∆ABC ≌ ∆ADC

∴ AD = BC, ∠CAD = ∠ACD and ∠ADC = ∠ABC. (Proved)

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 = 180^{0} —————————- (1)

And ∠ACB + ∠ACP = 180^{0} ————————- (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)