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Problem of the Week
Problem E and Solution
Slice of an Arc

Problem

Points \(A\), \(B\), and \(D\) lie on the circumference of a circle with centre \(C\).

If \(\angle ADB=x\degree\), then determine the measure of \(\angle ACB\) in terms of \(x\).

An illustration where the points A, B, and D all lie on the same half of the circle with centre C. Moving counterclockwise, the points are in the order A, B, D. Lines join A and B to C and join A and B to D.

Solution

We construct radius \(CD\). Let \(\angle CAD = p\degree\) and \(\angle CBD = q\degree\).

Since \(CA\) and \(CD\) are both radii of the circle, \(CA=CD\). So \(\triangle CAD\) is isosceles and \(\angle CDA=\angle CAD=p\degree\). Since the angles in a triangle add to \(180\degree\), \(\angle ACD=(180-2p)\degree\).

Since \(CB\) and \(CD\) are both radii of the circle, \(CB=CD\). So \(\triangle CBD\) is isosceles and \(\angle CDB=\angle CBD=q\degree\). Since the angles in a triangle add to \(180\degree\), \(\angle BCD=(180-2q)\degree\).

Now,

\[\begin{aligned} \hspace{2mm} \angle ACB&=\angle ACD-\angle BCD\\ &=(180-2p)\degree-(180-2q)\degree\\ &=(2q-2p)\degree\\ &=2(q-p)\degree \end{aligned}\]

Since \(\angle CDB = \angle CDA + \angle ADB\), we have \(q = p + x\).

Thus, \(\angle ACB = 2(q-p)\degree = 2x\degree\).

Note: In general, the angle inscribed at the centre of a circle is twice the size of the angle inscribed at the circumference by the same chord. That is, the angle inscribed by chord \(AB\) at the centre of the circle (\(\angle ACB\)) is double the angle inscribed by chord \(AB\) on the circumference (\(\angle ADB\)).