Tangent to Circles - Problem Solving

One tangent line, and only one, can be drawn to any point on the circumference of a circle, and this tangent is perpendicular to the radius through the point of contact. In the figure below, line \(BC\) is tangent to the circle at point \(A\).

If two tangents are drawn to a circle from an external point,

• the lengths of the tangents from the external point to the points of contact are equal;
• they subtend equal angles at the center of the circle.

In the figure below, \(AB=CB, \angle AOB= \angle COB\) and \(\angle ABO=\angle CBO\).

In the figure below, \(BA\) and \(BC\) are tangents to the circle at points \(A\) and \(C\), respectively. If \(\angle AEC=110^\circ\) and \(\angle DCE=80^\circ\), find \(\angle ABC\).

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SOLUTION

Consider the diagram to the right.

• Opposite angles of a cyclic quadrilateral sum up to \(180^\circ\). So \[\angle ADC=180^\circ-110^\circ=70^\circ.\]
• From the inscribed angle theorem, we can see that \[\angle AOC=2 \angle ADC=2(70^\circ)=140^\circ.\]

Since the sum of interior angle of a quadrilateral is \(360^\circ\), we have \[\begin{align} \angle ABC &=360^\circ-\angle BCO-\angle BAO-\angle AOC\\ &=360^\circ-90^\circ-90^\circ-140^\circ\\ &=40^\circ.\ _\square \end{align}\]