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:
  1. 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
     
  2. Since we extended BA to form line BD, ∠BAC + ∠DAC = 180°
    ⇒ y + ∠DAC = 180°
    ⇒ ∠DAC = 180° - y
  3. 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
     
  4. Now, ∠DCB = ∠DCA + ∠ACB =  
    180°-y
    2
      +  
    y
    2
      = 90°

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