1. ## Complex geometry

AB is the diameter and AC is a chord of a circle such that <BAC = 30o. If the tangent at C intersects AB produced in D, prove that BC = BD.

Is it even possible to prove it? I mean I think BC is less that half of BD...

2. Originally Posted by ice_syncer
AB is the diameter and AC is a chord of a circle such that <BAC = 30o. If the tangent at C intersects AB produced in D, prove that BC = BD.

Is it even possible to prove it? I mean I think BC is less that half of BD...
the picture is not accurate but it is something like that..

now, consider triangle $\Delta ABC$. what can you say about $\angle ACB$? take note the $A$ and $B$ are endpoints of diameter $AB$

3. Originally Posted by kalagota
the picture is not accurate but it is something like that..

now, consider triangle $\Delta ABC$. what can you say about $\angle ACB$? take note the $A$ and $B$ are endpoints of diameter $AB$
Sorry I didn't understand..

4. Originally Posted by ice_syncer
Sorry I didn't understand..
OK got it.

5. Originally Posted by ice_syncer
AB is the diameter and AC is a chord of a circle such that <BAC = 30o. If the tangent at C intersects AB produced in D, prove that BC = BD.

Is it even possible to prove it? I mean I think BC is less that half of BD...
angle BAC = 30 degrees ....given.

angle ACB = 90 degrees....because it is subtended by a diameter, by a semi-circle.

So, angle ABC = 180 -30 -90 = 60 degrees.

angle ABD is a straight angle, so it is 180 degrees.
And so, angle CBD = 180 -angle ABC = 180 -60 = 120 degrees.

Minor arc BC = 2(angle BAC) = 2(30) = 60 degrees....an inscribed angle is half of the intercepted arc in measure.
So, angle BCD = (1/2)(60) = 30 degrees....a tangent-chord angle is half of the intercepted arc in measure.
So, angle BDC = 180 -120 -30 = 30 degrees.

Hence, triangle CBD is an isosceles triangle, whose equal sides are BC and BD.

Therefore, BC = BD.

### a b is a diameter and ac is a chord of a circle such that angle BAC equals to 30 degrees is the tangent at C intersects

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