Question

please answer this question it's urgent

Answer 1

Sol: Given:- AB is the diameter with centre O, AC is the chord and CD is the tangent. Also given that ∠BAC= 30º. To prove:- BC=BD Proof:- In the fig., ∠ACB=90º [Angle in a semi-circle is 90º] ∠BAC=30º [Given] ∴ In the ΔABC, ∠ACB+∠BAC+∠ABC=180º [Sum of angles in a triangle] ⇒90º+30º+∠ABC=180º ⇒∠ABC=180º-(90º+30º) ⇒∠ABC=180º-120º ⇒∠ABC=60º ∴∠ABC=60º...............(I) Now, ∠BCD=∠BAC=30º [Angle on the alternate segment].................(II) Again, ∠ABC=∠BCD +∠BDC [Exterior angle is equal to sum of opposite interior angles] ⇒60º =30º+∠BDC [From (I) and (II)] ⇒∠BDC = 60º - 30º ⇒∠BDC = 30º ∴∠BDC =∠BCD =30º i.e. ∠BDC=∠BCD ⇒ BC=BD[Sides opposite to equal angles] Hence, BC=BD

Answer 2

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