Question

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

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