Question

in the given figure Δ ABC is an isoceles triangle in which angle AB=AC . Side BA is produced to D such that AD=AC. show that angle BCD is a right angle.

Answer 1

in triangle ABC, AB = AC is given
angle ABC= angle ACB (opposite angles are equal in triangle)
in another triangle AD=AC
angle ACD= angle ADC (opposite angles are equal in triangle)
In triangle BCD,
 angle ABC + angle BCD + angle ADC = 180 degrees
angle ACB+angle ACB+angle ACD+angle ACD = 180 degrees
2 (angle ACB+angle ACD) = 180 degrees
2(angle BCD) = 180 degrees
angle BCD is 180/2 = 90 degrees.
Hence, proved.

Answer 2

Hello mate ^_^

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$\bold\pink{Solution:}$

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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