Question 6:
∆ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that
AD = AB (see the given figure). Show that ∠BCD is a right angle.
Answer:
In ∆ABC,
AB = AC (Given)
∠ACB = ∠ABC (Angles opposite to equal sides of a triangle are also equal)
In ∆ACD,
AC = AD
∠ADC = ∠ACD (Angles opposite to equal sides of a triangle are also equal)
In ∆BCD,
∠ABC + ∠BCD + ∠ADC = 180º (Angle sum property of a triangle)
∠ACB + ∠ACB +∠ACD + ∠ACD = 180º
2(∠ACB + ∠ACD) = 180º
2(∠BCD) = 180º
∠BCD = 90º
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