(This problem is from CE Ref V4 of Gillesania)

A quadrilateral ABCD is inscribed in a semicircle with side AD as the diameter. If point O is the center of the diameter, determine the angle DCO if angle CAD is 39 deg.

In the book's solution, it says that angle COD = 2 times angle CAD. Is this a rule in semi circles?

(Attached is the drawn image of the problem)

Thanks in advance sirs.

https://drive.google.com/open?id=0BzglifbmrqfKOUtqSFZEWnNYNVE

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yes. Not only for semicircles but for all circles. Try searching for inscribed angles.

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CAO is an isosceles triangle, because AO and CO are radii of the circle.
Angles CAD=CAO and OCA are the base angles in isosceles triangle CAO,
both with the same measure, because it is an isosceles triangle.
Angle COD is an exterior angle for triangle CAO, and an exterior angle 's measure is the sum of the measures of the two (interior) angles of the triangle that are not adjacent.
Those two non-adjacent interior angles are CAD and OCA, and both have the same measure,
so the measure of COD is twice the measure of CAD.

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