In the figure below, if the radius of circle O is r and tria

Bunuel - Thanks. But how can (angle DOC is 60°)?

DOC is 30°-60°-90° right triangle. Angle D=90°, angle DCO=30° and angle DOC=60°.

Sorry Bunuel - what I meant to ask was - How come OC will bisect the angle DCO?

Ask yourself, why should any angle from ACO and BCO be greater than another?

Bumping for review and further discussion.

Hi Bunuel, can we say that in equilateral triangle median = angle bisector = altitude = perpendicular bisector?

Yes, that's true. For more check here: math-triangles-87197.html

Hi Bunnel,

How we solve this problem by using the direct formula for inradius of equilateral triangle

how does the line bisect the angle 60?

thanks

It a special property of incircles of equilateral triangles. Look below for the proof.

Triangle ABC is an equilateral traingle with O as the center of the incircle. Radii of the incircle (OD and OF) are perpendicular to the sides AC and BC (by property of a circle wherein radius is perpendicular to the tangent at the point of tangency, CD and CF are tangents to the incircle at points of tangencies D and F respectively).

Additionally, for a circle, 2 tangents (CD and CF) drawn from the same external point (C) are of same length --> CD = CF

Thus, in right triangles CDO and CFO,

CD=CF
OD=OF = radius of the incircle
OC=OC (common side)

Thus Triangle CDO is congruent to triangle CFO (by SSS property of congruence)--->\(\angle { DCO} = \angle{FCO}\) and thus \(\angle { DCO} = \angle{FCO} = \frac{\angle {ACB}}{2}\)

Dont need to learn the proof but do remember that the line drawn from the center of an incircle of an equilateral triangle to one of the veritces will always bisect that angle.

Hope this helps.

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