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Home » AB is 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 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.
Class 10 | Exercise 18B | ICSE | Maths | Selina | Tangents and Intersecting Chords
AB is 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.

Selina Solutions Concise Class 10 Maths Chapter 18 ex. 18(B) - 7

Join

    \[OC\]

    \[\angle BCD\text{ }=\angle BAC\text{ }=\]

    \[{{30}^{o}}~\]

[Angles in the alternate segment]

It’s seen that, arc

    \[BC\]

subtends

    \[\angle DOC\]

at the center of the circle

And

    \[\angle BAC\]

at the remaining part of the circle.

So,

    \[\angle BOC\text{ }=\text{ }2\angle BAC\text{ }=\text{ }2\text{ }x\text{ }{{30}^{o}}~=\text{ }{{60}^{o}}\]

Now, in

    \[\vartriangle OCD\]

    \[\angle BOC\text{ }or\angle DOC\text{ }=\text{ }{{60}^{o}}\]

    \[\angle OCD\text{ }=\text{ }{{90}^{o}}~[OC\bot CD]\]

Or,

    \[\angle DOC\text{ }+\angle ODC\text{ }=\text{ }{{90}^{o}}\]

    \[\angle ODC\text{ }=\text{ }{{90}^{o}}-\text{ }{{60}^{o}}~=\text{ }{{30}^{o}}\]

Now, in

    \[\vartriangle BCD\]

As

    \[\angle ODC\text{ }or\angle BDC\text{ }=\angle BCD\text{ }=\text{ }{{30}^{o}}\]

Therefore,

    \[BC\text{ }=\text{ }BD\]

i

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