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'
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Hello mate ^_^
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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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