Question

AC is a diameter of •E and BC || ED. Find the measure of arc CD.

Answer 1

Based on the image provided, it is evident that \(BC \parallel ED\). Additionally, it is given that \(AC\) is the diameter. Consequently, we can deduce that the radius (\(BE = ED = CE = AE\)).

Considering triangle \(BCE\), the sum of angles \(\angle BEC\) and \(\angle BEA\) equals 180° (due to the angle on a straight line property, \(\angle BEC + \angle BEA = 180°\)). Simplifying, \(180° - 90° = \angle BEC = 90°\), establishing that triangle \(BCE\) is an isosceles triangle with equal sides \(BE\) and \(CE\).

As \(BC\) is parallel to \(ED\), we can conclude that \(\angle BCE = \angle CED\) (alternate angles are equal), which equals \(45°\). Therefore, the final answer for arc CD is \(arcCD = 45°\).

Answer 2

From the image in the question, we have that

`BC\parallel ED`

Given also

`AC=diameter`

From the given information above, we will have that

`radius=BE=ED=CE=AE`

From the triangle BCE,

`\begin{gathered} \angle BEC+\angle BEA=180^0(angle\text{ on straight line} \\ \angle BEC+90^0=180^0 \\ \angle BEC=180^0-90^0 \\ \angle BEC=90^0 \end{gathered}`

Since the triangle BEC is bounded by two radii CE and BE,

Then we can say that the triangle BEC is an Isosceles triangle

A triangle with two sides of equal length is an isosceles triangle. The two equal sides of an isosceles triangle are known as 'legs' whereas the third or unequal side is known as the 'base'. Many things in the world have the shape of an isosceles triangle.

Two angles are also equal and they are given below as

`\angle CBE=\angle BCE=45^0`

Since,

BC is parallel to ED, we will have the final answer as

`\begin{gathered} \angle BCE=\angle CED(alternate\text{ angles are equal\rparen} \\ \angle CED=45^0 \end{gathered}`

Therefore,

The final answer for arc CD is

`\Rightarrow arcCD=45^0`