in isosceles triangle ABC AB=AC. The side BA is produced to D such that BA=BD.Prove that angle BCD=90 degree

Given in ΔABC, AB = AC
⇒ ∠ABC = ∠ACB (Since angles opposite to equal sides are equal)
Also given that AD = AB
⇒ ∠ADC = ∠ACD (Since angles opposite to equal sides are equal)
∴ ∠ABC = ∠ACB = ∠ADC = ∠ACD = x (AB = AC = AD)
In ΔBCD, ∠B + ∠C + ∠D = 180°
x + 2x + x = 180°
4x = 180°
x = 45°
∠C =  2x = 90°
Thus BCD is a right angled triangle

Ok, so

 Given : AB = AC. , AB = AD 

            To Prove : BCD = 90. 

 

Proof: In triangle ABC , AB = AC ( Given )......(1)

           AB = AD ( Given ) …......(2)

From (1) and (2), AC = AD. 

 

Now in Triangle BCD, since AD= AC, ADC = ACD (Angles opposite to equal sides are equal)

 

In triangle ABC, ABC = ACB ( Since AB = AC, angles opposite to equal sides are equal )

 

Mark ACB and ABC as “x” and ADC and ACD as “y” ( Since both of these pairs of angles are equal ) ….........{3}

 

   Now in triangle BCD, By Angle Sum Property of a traingle , x +x+y+y = 180.  (from {4} )

 

2 (x + y) = 180.

 

Therefore x + y 90.

 

But remember that x + y is actually angle BCD ??

 

Therefore BDC = 90.

 

Hence Proved.

 

Hope this answer helps you.

AB = AC
Therefore we can say that ∠ABC = ∠ACB ( angles opposite to equal sides are equal)
Also given that AD = AB
Then we can aslo say that ∠ADC = ∠ACD (Since angles opposite to equal sides are equal)