triangle ABC is isosceles triangle with ab is equal to AC side BA reduce D that AB is equal to AC prove that angle abc is a right angle
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Given: ∆ABC is an isosceles ∆.
AB = AC and AD = AB
To Prove:
∠BCD is a right angle.
Proof:
In ΔABC,
AB = AC (Given)
⇒ ∠ACB = ∠ABC (Angles opposite to the equal sides are equal.)
In ΔACD,
AD = AB
⇒ ∠ADC = ∠ACD (Angles opposite to the equal sides are equal.)
Now,
In ΔABC,
∠CAB + ∠ACB + ∠ABC = 180°
⇒ ∠CAB + 2∠ACB = 180°
⇒ ∠CAB = 180° – 2∠ACB — (i)
Similarly in ΔADC,
∠CAD = 180° – 2∠ACD — (ii)
also,
∠CAB + ∠CAD = 180° (BD is a straight line.)
Adding (i) and (ii)
∠CAB + ∠CAD = 180° – 2∠ACB + 180° – 2∠ACD
⇒ 180° = 360° – 2∠ACB – 2∠ACD
⇒ 2∠ACB + 2∠ACD= 360-180
⇒ 2(∠ACB + ∠ACD) = 180°
⇒ ∠BCD = 90°
AB = AC and AD = AB
To Prove:
∠BCD is a right angle.
Proof:
In ΔABC,
AB = AC (Given)
⇒ ∠ACB = ∠ABC (Angles opposite to the equal sides are equal.)
In ΔACD,
AD = AB
⇒ ∠ADC = ∠ACD (Angles opposite to the equal sides are equal.)
Now,
In ΔABC,
∠CAB + ∠ACB + ∠ABC = 180°
⇒ ∠CAB + 2∠ACB = 180°
⇒ ∠CAB = 180° – 2∠ACB — (i)
Similarly in ΔADC,
∠CAD = 180° – 2∠ACD — (ii)
also,
∠CAB + ∠CAD = 180° (BD is a straight line.)
Adding (i) and (ii)
∠CAB + ∠CAD = 180° – 2∠ACB + 180° – 2∠ACD
⇒ 180° = 360° – 2∠ACB – 2∠ACD
⇒ 2∠ACB + 2∠ACD= 360-180
⇒ 2(∠ACB + ∠ACD) = 180°
⇒ ∠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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