[tex]\large\underline{\sf{Solution-1}}[/tex]

Given that,

∆ABC is an isosceles triangle in which AB = AC.

[tex]\sf \:  \implies \: \angle \: ABC = \angle \: ACB \: = \: x \: (say) - - - (1) \\ \\ [/tex]

[ Angle opposite to equal sides are equal ]

Further given that,

Side BA is produced to D such that AD is equal to AB.

Now, AD = AB = AC

[tex]\sf \:  \implies \: \angle \: ADC = \angle \: ACD \: = \: y \: (say) - - - (2) \\ \\ [/tex]

[ Angle opposite to equal sides are equal ]

Now, In triangle BCD

We know, sum of all interior angles of a triangle is 180°.

So,

∠DBC + ∠BCD + ∠CDB = 180°

[tex]\sf \: x + x + y + y = 180 \degree \\ \\ [/tex]

[tex]\sf \: 2(x + y) = 180 \degree \\ \\ [/tex]

[tex]\sf \:\sf \:  \implies \: x + y = 90 \degree \\ \\ [/tex]

[tex]\sf \:\bf \:  \implies \: \angle \: BCD = 90 \degree \\ \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

[tex]\large\underline{\sf{Solution-2}}[/tex]

Given that,

ABC is a right angled triangle in which ∠A = 90° and AB = AC.

[tex]\sf \:  \implies \: \angle \: ABC = \angle \: ACB \: = \: x \: (say) \\ \\ [/tex]

Now, In right-angle triangle ABC,

[tex]\sf \: \angle \: ABC + \angle \: ACB \: + \angle \: BAC \: = \: 180 \degree \: \\ \\ [/tex]

[tex]\sf \: x + x + 90\degree = \: 180 \degree \: \\ \\ [/tex]

[tex]\sf \: 2x = 180\degree - 90\degree \\ \\ [/tex]

[tex]\sf \: 2x = 90\degree \\ \\ [/tex]

[tex]\sf \:  \implies \: x = 45\degree \\ \\ [/tex]

So,

[tex]\bf \:  \implies \: \angle \: ABC = \angle \: ACB \: = \: 45\degree \: \\ \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

[tex]\large\underline{\sf{Solution-3}}[/tex]

Let assume that ABC be an equilateral triangle.

As, triangle ABC is an equilateral triangle.

So, AB = BC = AC

Now, AB = AC

[tex]\sf \:  \implies \: \angle \: ABC = \angle \: ACB \: - - - (1) \\ \\ [/tex]

Again, AB = BC

[tex]\sf \:  \implies \: \angle \: BAC = \angle \: ACB - - - (2) \\ \\ [/tex]

From equation (1) and (2), we concluded that

[tex]\sf \:  \implies \: \angle \: ABC = \angle \: ACB \: = \angle \: BAC \: = \: x \: say\\ \\ [/tex]

Now, In triangle ABC,

[tex]\sf \: \angle \: ABC + \angle \: ACB \: + \angle \: BAC \: = \: 180 \degree \: \\ \\ [/tex]

[tex]\sf \: x + x + x = 180\degree \\ \\ [/tex]

[tex]\sf \: 3x = 180\degree \\ \\ [/tex]

[tex]\sf \:  \implies \: x = 60\degree \\ \\ [/tex]

So,

[tex]\bf \:  \implies \: \angle \: ABC = \angle \: ACB \: = \angle \: BAC \: = \: 60\degree\\ \\ [/tex]

Answer:

Hello User!!

1)

Given that:-

• AB = AC and AD = AB

To find:-

• ∠BCD is a right angle.

Solution:-

In ∆ABC,

AB=AC

∠ACB=∠ABC (i)

In ∆ACD,

AC=AD

∠ACD=∠ADC (ii)

On adding equations (i) and (ii) we get

∠ACB+∠ACD=∠ABC+∠ADC

∠BCD=∠ABC+∠BDC

Adding ∠BCD on both side

∠BCD+∠BCD=∠ABC+∠BDC+∠BCD

2∠BCD=180°

[tex] \angle \: bcd = \frac{ { \cancel{180}} \: ^{90} }{ \cancel2} \\ [/tex]

= 90°

2)

Given that:-

• AB = AC

To find:-

• ∠B and ∠C.

Solution:-

It is given that,

AB = AC

∴ ∠C = ∠B (Angles opposite to equal sides are also equal)

Let ∠B = ∠C = x

In ΔABC,

∠A + ∠B + ∠C = 180° (Angle sum property of a triangle)

[tex]90 \degree + x + x = 180 \degree \\ \\ 90 \degree + 2x = 180 \degree \\ \\ 2x = 90 \degree \\ \\ \therefore \: x = \frac{ { \cancel{90}} \: ^{45} }{ \cancel2} \\ \\ \sf = 45 \degree[/tex]

∠B = ∠C = 45°

3)

Drawing am equilateral Triangle-in attachment,

Therefore,

AB = BC = AC

∴ ∠C = ∠A = ∠B (Angles opposite to equal sides of a triangle are equal)

Let ∠A = ∠B = ∠C be x.

In △ ABC,

∠A + ∠B + ∠C = 180° (Angle sum property of a triangle)

[tex]x + x + x = 180 \degree[/tex]

[tex] \implies3x = {180 \degree} \\ \\ \therefore \: x = \frac{ { \cancel{180}} \: ^{60} }{ \cancel3} \\ \\ x = 60 \degree [/tex]

Hence, in an equilateral triangle, all interior angles are of measure 60°.

Hope it helps you from my side!!

Keep Learning and solving!!