∆ABC is an isosceles triangle with …
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Gaurav Seth 5 years, 3 months ago
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.
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