AB is the diameter and AC is a chord of a circle with centre O such that angle BAC = 30o. The tangent to the circle at C intersects AB produced in D. Show that : BC = BD.
AB is the diameter and AC is a chord of a circle with centre O such that angle BAC = 30o. The tangent to the circle at C intersects AB produced in D. Show that : BC = BD.
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
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
