AB is a diameter of a circle and AC is its chord such that `angleBAC=30^(@)`. If the tangent at C intersects AB extended at D, then BC=BD.

Explanation

A circle is drawn with center O and AB is a diameter
AC is a chrod such that `angleBAC=30^(@)`.
Given AB is `prop` diameter and AC is a chord of circle with centre O, `angleBAC=30^(@).`
To prove BC=BD

Proof `angleBCD=angleCAB` [alternate segment theorem]
`angleCAB=30^(@)` [given]
`:.angleBCD=30^(@).....(i)`
In `DeltaABC, angleACB=90^(@)` [angle in semi-circle is right angle]
`angleA+angleB+angleC=180^(@)`
`30^(@)+angleB+90^(@)=180^(@)`
`rArrangleB=60^(@)`
Also, `angleCBA+angleCBD=180^(@)` [linear pair]
`rArrangleCBD=180^(@)-60^(@)-120^(@)` `[:'angleCBA=60^(@)]`
Now, in `DeltaCBD`
`angleCBD+angleBDC+angleDCB=180^(@)`
`rArr120^(@)+angleBDC+30^(@)=180^(@)` [from Eq. (i)]
`rArrangleBDC=30^(@) rArrangleBDC=30^(@).....(ii)`
From Eqs. (i) and (ii),
`angleBCD=angleBDC`
`:.BC=BD`
[sides opposite to equal angles are equal]