ABCD is an isoceles triangle in which AB = AC. side BA is produced to D such that AD=AB. Show that BCD is a right angle.
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Given in ΔABC, AB = AC
⇒ ∠ABC = ∠ACB (Since angles opposite to equal sides are equal)
Also given that AD = AB
⇒ ∠ADC = ∠ACD (Since angles opposite to equal sides are equal)
∴ ∠ABC = ∠ACB = ∠ADC = ∠ACD = x (AB = AC = AD)
In ΔBCD, ∠B + ∠C + ∠D = 180°
x + 2x + x = 180°
4x = 180°
x = 45°
∠C = 2x = 90°
Thus BCD is a right angled triangle
⇒ ∠ABC = ∠ACB (Since angles opposite to equal sides are equal)
Also given that AD = AB
⇒ ∠ADC = ∠ACD (Since angles opposite to equal sides are equal)
∴ ∠ABC = ∠ACB = ∠ADC = ∠ACD = x (AB = AC = AD)
In ΔBCD, ∠B + ∠C + ∠D = 180°
x + 2x + x = 180°
4x = 180°
x = 45°
∠C = 2x = 90°
Thus BCD is a right angled triangle
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RIGHT ANGLE IS C =90
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I THINK THE ANSWER IS CORRECT
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