Question

Triangle abc is an issocles triangle with AB=AC ,side BA is produced to D such that AB=AD. Prove that angle BCD is a right angle triangle

Answer 1

In ∆ABC, we have,          AB = AC  [given]⇒∠ACB = ∠ABC   [angles opposite to equal sides are equal] .......(1)In ∆ACD, we have,          AD = AC  [given]⇒∠ACD = ∠ADC   [angles opposite to equal sides are equal] ........(2)In ∆ ADC, we have    ∠DBC + ∠BCD + ∠BDC = 180° [Angle sum property]⇒∠ABC + ∠BCD + ∠ADC = 180°⇒∠ACB +∠BCD + ∠ACD =  180°  [using (1) and (2)]⇒∠BCD + (∠ACB+∠ACD) = 180°⇒∠BCD + ∠BCD = 180°⇒2∠BCD =180°⇒∠BCD = 90°
 

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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