ΔABC is an isosceles triangle in which AB=AC side BA is produced to D such that AD=AB show that ∠BCD=90°
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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 )
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 )
Answered by
2
Hello mate ^_^
__________________________/\_
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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