Question

triangle ABC is isoceles with AB=AC, BA is produced to D such that AB=AD. Prove that Angle BCD=90

Answer 1

given,  AB = AC
Also, AB = AD          hence,  AD= AC

In the triangle ABC, angle B = angle ACB
In the triangle ACD, angle ACD = angle C - angle B

In the triangle ACD,  angle D = angle ACD  as  AD = AC

=>  angle C - angle B = angle D
=>  angle C = angle B + angle D

since sum of angles B, C and D is 180 deg.
      angle C = 90 and angle B + angle D = 90.
  so angle BCD =90 deg.

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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