Question

triangle ABC is an isosceles triangle in which AB = AC side BA is produced to D such that AD=AB show that angle BCD is a right angle

Answer 1

It is very easy
In triangle ABC
AB = AC   (Given)
So, angle ABC = angle ACB = x  (In an isosceles triangle opposite angles                                                              are equal)
In triangle ACD
AC = AD   (Given)
So, angle ACD = angle ADC = y
 
In triangle DBC
angle DBC + angle BCD + angle CDB = 180
x + (x+y) + y = 180
x + x + y + y =180
2x + 2y = 180
2(x+y) = 180
x+y = 180/2
x+y = 90

Angle BCD = angle (x+y) = 90

Answer 2

Hello mate ^_^

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$\bold\green{Solution:}$

AB=AC         (Given)

It means that ∠DBC=∠ACB           (In triangle, angles opposite to equal sides are equal)     

Let ∠DBC=∠ACB=x         .......(1)

AC=AD          (Given)

It means that ∠ACD=∠BDC         (In triangle, angles opposite to equal sides are equal)     

Let ∠ACD=∠BDC=y           ......(2)

In ∆BDC, we have

∠BDC+∠BCD+∠DBC=180°     (Angle sum property of triangle)

⇒∠BDC+∠ACB+∠ACD+∠DBC=180°

Putting (1) and (2) in the above equation, we get

y+x+y+x=180°

⇒2x+2y=180°

⇒2(x+y)=180°

⇒(x+y)=180/2=90°

Therefore, ∠BCD=90°

hope, this will help you.☺

Thank you______❤

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