Question

ΔABC is an isosceles triangle in which AB=AC side BA is produced to D such that AD=AB show that ∠BCD=90°

Answer 1

see,in ΔABC
 AB = AC
⇒∠ACB = ∠ABC ( opp. angles of equal sides)
also,in ΔACD
AC= AD
∵AB = AD
so, ∠ADC = ∠ACD ( opp. angles of equal sides)
in ΔBCD
∠ABC + ∠ACB +∠ACD + ∠ADC = 180° ( sum of all angles in a triangle)
∵∠ACB +∠ACD  = ∠BCD
⇒2∠ACB =  ∠BCD
4∠ACB = 180°
⇒∠ACB = 45°
⇒2∠ACB  = 90°
⇒∠BCD = 90° (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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