Problem – 5: In the figure, AD = AE, BD = CE and ∠AEC = ∠ADB. Prove that AB = AC.

Particular enunciation: In the figure, AD = AE, BD = CE and ∠AEC = ∠ADB. We have to prove that AB = AC.

Proof: In ∆ACE and ∆ADB,

AD = AE [Given]

BD = CE [Given]

and ∠AEC = ∠ADB

∴ ∆ACE ≌ ∆ADB

Therefore, AB = AC. (Proved)

Problem – 6: In the figure, ∆ABC and ∆DBC are both isosceles triangles. Prove that, ∆ABD ≌ ∆ACD.

Particular enunciation: In the figure, ∆ABC and ∆DBC are both isosceles triangles. We have to prove that, ∆ABD ≌ ∆ACD.

Proof: Given that, ∆ABC is an isosceles triangle.

∴ AB = AC

And ∆DBC is an isosceles triangle.

∴ DB = DC

Now, in ∆ABC and ∆ACD

AB =AC

DB = DC

And AD = AD [common side]

∴ ∆ABD ≌ ∆ACD (Proved)

Problem – 7: Show that the medians drawn from the extremities of the base of an isosceles triangle to the opposite sides are equal to each other.

General enunciation: The medians drawn from the extremities of the base of an isosceles triangle to the opposite sides are equal to each other.

Particular enunciation: Let, ABC is an isosceles triangle where AB = AC. BQ and CP are medians drawn to the sides AC and AB respectively. We have to prove that, BQ = CP.

Proof: In ∆ABC,

AB =AC

⟹ ½ AB = ½ AC [Divide both side by 2]

⟹ BP = CQ [P and Q are the mid points of AB and AC respectively]

Now in ∆BCE and ∆DCB,

BP = CQ

∴ ∠PBC = ∠QCB [∵ opposite angles of equal arms AB and AC are equal]

And BC is the common side.

∴ ∆BCP ≌ ∆QCB

∴ BQ = CP [proved]

Problem – 8: Prove that the angles of an equilateral triangle are equal to one another.

General enunciation: We have to prove that the angles of an equilateral triangle are equal to one another.

Particular enunciation: Let, ABC is an equilateral triangle i.e., AB = AC = BC.

We have to prove that, ∠A = ∠B = ∠C

Proof: Given that, AB =AC = BC

In ∆ABC,

AB = AC

∴ ∠ACB = ∠BAC [∵opposite angles are equal arms are equal.]

⟹∠C = ∠B —————————————- (1)

Again, in ∆ABC,

AC = BC

∴ ∠ABC = ∠BAC [∵opposite angles are equal arms are equal.]

⟹ ∠B = ∠A ————————————— (2)

From (1) and (2) we have

∠C = ∠B = ∠A

∴ ∠A = ∠B = ∠C [proved]