### ABC is an isosceles triangle with AB = AC. We extend the segment BA to D such that BA = AD. What is the value of ∠BCD?

**Answer:**

90°

**Step by Step Explanation:**

- Take a look at the image below:

ABC is an isosceles triangle with AB = AC,

therefore, ∠ABC = ∠ACB = x

Let ∠CAB = y. Then y + x + x = 180°

⇒ y + 2x = 180°

⇒ ∠ACB = x =180°-y 2 - Since we extended BA to form line BD, ∠BAC + ∠DAC = 180°

⇒ y + ∠DAC = 180°

⇒ ∠DAC = 180° - y - CAD is also an isosceles triangle since AD = AC (remember AD = AB and AB = AC)

So ∠CDA = ∠DCA = z

⇒ ∠CDA + ∠DCA + ∠DAC = 180°

⇒ z + z + (180°-y) = 180°

⇒ 2z = y

⇒ ∠DCA = z =y 2 - Now, ∠DCB = ∠DCA + ∠ACB =

+180°-y 2

= 90°y 2