Question

in the given figure ABC is a triangle in which AB = AC side BA is produced to D such that AB=AD.prove that angle BCD= 90 degree

Answer 1

in triangle ABC ,
AB=AC
=> angle ABC = angle ACB
Let angle ABC = angle ACB = x
Similarly in triangle ACD ,
AC = AD
=> angle ACD = angle ADC
Let angle ACD = angle ADC = y
in triangle BCD ,
angle DBC + angle BCD + angle CDB = 180°
=> ang ABC + ang ACB + ang ACD + ang ADC = 180°
=> x + x + y + y = 180°
=> 2×( x + y ) = 180°
=> x + y = 90°
=> ang ACB + ang ACD = 90°
=> angle BCD = 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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