Find Angle DCO: Semicircle Geometry with Equal Radii

Grades 9-10

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Question: AO=DC, find angle DCO

To find the measure of

\angle DCO

, we can use the properties of circles and triangles. Let's break down the geometric relationships step by step:

Identify Equal Radii: Since $O$ is the center of the semicircle, the segments connecting the center to any point on the arc are equal. Thus, the radii $OA$ , $OE$ , and $OD$ are all equal in length.
Utilize the Given Condition: The problem states that $AO = DC$ . Since we already established that $AO$ is a radius, it follows that the segment $DC$ is also equal to the radius of the semicircle.
Analyze Triangle $ODC$ : In $\triangle ODC$ , we now know that $OD = DC$ (both are equal to the radius). This makes $\triangle ODC$ an isosceles triangle. Let the angle we are looking for, $\angle DCO$ , be denoted as $x$ . In an isosceles triangle, the angles opposite the equal sides are equal, so $\angle DOC = \angle DCO = x$ .
Find the Exterior Angle: Looking at $\triangle ODC$ , the angle $\angle EDO$ is an exterior angle at vertex $D$ . By the exterior angle theorem, an exterior angle of a triangle is equal to the sum of the two opposite interior angles. Therefore, $\angle EDO = \angle DOC + \angle DCO = x + x = 2x$ .
Analyze Triangle $OED$ : Since $OE$ and $OD$ are both radii, $\triangle OED$ is also an isosceles triangle. This means the base angles are equal: $\angle OED = \angle EDO = 2x$ .
Relate to the Given $45^\circ$ Angle: Now consider the large triangle $\triangle OEC$ . The line $AOC$ is straight, so the angle $\angle EOA = 45^\circ$ is an exterior angle to $\triangle OEC$ at vertex $O$ . Using the exterior angle theorem again: $\begin{align*} \angle EOA &= \angle OEC + \angle ECO \\ 45^\circ &= 2x + x \\ 45^\circ &= 3x \\ x &= 15^\circ \end{align*}$

Thus, the measure of

\angle DCO

is $15^\circ$ .