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ABC is an isosceles triangle in which AB = ACSide BA is produced to D such that AD = AB(see Fig. 7.34). Show that 2 BCD is a right angle |
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Answer» ∆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° ========================================= |
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