Given: ∆ABC is an isosceles triangle in which AB = AC.
Side BA is produced to D such that AD = AB.
To Prove: ∠BCD is a right angle.
Proof: ∵ ABC is an isosceles triangle
∴ ∠ABC = ∠ACB    ...(1)
∵ AB = AC and AD = AB
∴ AC = AD

∴ In ∆ACD,
∠CDA = ∠ACD
| Angles opposite to equal sides of a triangle are equal
⇒ ∠CDB = ∠ACD    ...(2)
Adding the corresponding sides of (1) and (2), we get
∠ABC + ∠CDB = ∠ACB + ∠ACD
⇒ ∠ABC + ∠CDB = ∠BCD    ...(3)
In ∆BCD,
∠BCD + ∠DBC + ∠CDB = 180°
| ∵ Sum of all the angles of a triangle is 180°
⇒ ∠BCD + ∠ABC + ∠CDB = 180°
⇒    ∠BCD + ∠BCD = 180°
| Using (3)
⇒    2∠BCD = 180°
⇒    ∠BCD = 90°
⇒ ∠BCD is a right angle.

Given: ABC is an isosceles triangle in which altitudes BE and CF are drawn to sides AC and AB respectively.
To Prove: BE = CF.
Proof: ∵ ABC is an isosceles triangle
∴ AB = AC
∴ ∠ABC = ∠ACB    ...(1)
| Angles opposite to equal sides of a triangle are equal
In ∆BEC and ∆CFB,
∠BEC = ∠CFB    | Each = 90°
BC = CB    | Common
∠ECB = ∠FBC    | From (1)
∴ ∆BEC ≅ ∆CFB    | By AAS Rule
∴ BE = CF.    | C.P.C.T.

Given: ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal.
To Prove: (i) ∆ABE = ∆ACF
(ii) AB = AC, i.e., ∆ABC is an isosceles triangle.
Proof: (i) In ∆ABE and ∆ACF
BE = CF    | Given
∠BAE = ∠CAF    | Common
∠AEB = ∠AFC    | Each = 90°
∴ ∆ABE ≅ ∆ACF    | By AAS Rule
(ii) ∆ABE ≅ ∆ACF | Proved in (i) above
∴ AB = AC    | C.P.C.T.
∴ ∆ABC is an isosceles triangle.

∵ In ∆ABC,
AB = AC
∴ ∠B = ∠C    ...(1)
| Angles opposite to equal sides of a triangle are equal
In ∆ABC,
∠A + ∠B + ∠C = 180°
| Sum of all the angles of a triangle is 180°
⇒ 90° + ∠B + ∠C = 180°
| ∵ ∠A = 90° (given)
⇒    ∠B + ∠C = 90°    ...(2)
From (1) and (2), we get
∠B = ∠C = 45°.

Given: ABC and DBC are two isosceles triangles on the same base BC.
To Prove: ∠ABD = ∠ACD.
Proof: ∵ ABC is an isosceles triangle on the base BC.
∴ ∠BC = ∠ACB    ...(1)
∵ DBC is an isosceles triangle on the base BC
∴ ∠DBC = ∠DCB    ...(2)
Adding the corresponding sides of (1) and (2), we get
∠ABC + ∠DBC = ∠ACB + ∠DCB
⇒    ∠ABD = ∠ACD.