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AB is a diameter of a circle and AC is i...

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

Text Solution

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To Prove, BC=BD

Join BC and OC.
Given, `angleBAC=30^(@)`
`rArrangleBCD=30^(@)`
[angle between tangent and chord is equal to angle made by chord in the alternate segment]
`:.angleACD=angleACO+angleOCD=30^(@)+90^(@)=120^(@)`
`[:.OCbotCDandOA=OC=radiusrArrangleOAC=angleOCA=30^(@)]`
In `DeltaACD,` `angleCAD+angleACD+angleADC=180^(@)`
[since, sum of all interior angles of a triangle is `180^(@)`]
`rArr30^(@)+120^(@)+angleADC=180^(@)`
`rArrangleADC=180^(@)-(30^(@)+120^(@))=30^(@)`
Now in `DeltaBCD` `angleBCD=angleBDC=30^(@)`
`rArrBC=BD`
[since, sides opposite to equal angles are equal]
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