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

True.
CD is a tangent at contact point C. AOB is diameter which meets tangent produced at D.
Chord AC makes `angleA=30^(@)` with diameter AB.
To Prove : `BD-BC`
Proof : In `triangleOAC,`
`OA=OC=r" "`(radii of same dircle)
`angle1=angleA" "`(angles opposite to equal sides are equal)
`implies" "angle1=30^(@)" " (because angleA=30^(@))`
Exterior `angleBOC=angle2=angle1+angleA=(30^(@)+30^(@))=60^(@)`
Now, in `triangleOCB,`
`OB=OC" "`(radii of same circle)
`:." "angle3=angle4" "`(angles opposite to equal sides are equal)
`angle3+angle4+angleCOB=180^(@)`
`implies" "angle3+angle3+60^(@)=180^(@)" "`(angle sum property of triangle)
`implies" " 2angle3=180^(@)-60^(@)=120^(@)`
`implies" "angle3=60^(@)=angle4`
`implies" "angle6+angle4=180^(@)" "`(linear pair axiom)
`implies" "angle6=180^(@)-angle4=180^(@)-60^(@)`
`implies" "angle6=120^(@)`
`:.` Tangent CD and radius CO are at contact point C.
`:." "angleOCD=90^(@)`
(radius through point of contact is perpendicular to the tangent )
`implies" "angle3+angle5=90^(@)" "implies" "60^(@)+angle5=90^(@)`
`implies" "angle5=30^(@)`
Now, in `triangleBCD,`
`:." "angle5+angle6+angleD=180^(@)" "`(angle sum property)
`implies" "120^(@)+30^(@)+angleD=180^(@)" "implies" "angleD=30^(@)`
Since, `angleC=angleD=30^(@).`
Hence, `BC=BD" "`(sides opposite to equal angles)
Hence Proved.