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 $\begin{aligned} \angle ABC &=360^\circ-\angle BCO-\angle BAO-\angle AOC\\ &=360^\circ-90^\circ-90^\circ-140^\circ\\ &=40^\circ.\ _\square \end{aligned}$