Constructing a Tangent Line to a Circle from a Point Outside Circle

We also need to know how to construct a tangent line to a circle from a point outside the circle.

Steps for Constructing a Tangent to a Circle Through a Point Outside the Circle

Start with any circle (label the center O) and any point outside the circle (point P).
Connect point O to point P.
Construct the perpendicular bisector of segment OP.
Construct an arc using the midpoint of OP as the center and half the length of OP as the radius.
Label the points where the arc intersects the circle.
Construct rays from P through the points of intersection.
Both rays you have constructed are tangent to the circle!

There is one more important shape we need to talk about. This shape is the cyclic quadrilateral. A cyclic quadrilateral is a quadrilateral inscribed in a circle –that is, all 4 vertices lie on the circle. Cyclic quadrilaterals have some special properties.

Circle with lines and points The sum of opposite angles of a cyclic quadrilateral is 180 degrees. Let's prove this! We will use the diagram to the left for reference.

To show that the sum of opposite angles of a cyclic quadrilateral is 180 degrees, we need to prove:

$LaTeX: \angle DAB+\angle DCB=\angle ABC+ADC=180^\circ$

Proof

We start by drawing segments DO and BO.

By the Inscribed Angles Theorem, we know that equation image indicator $LaTeX: \angle DAB=\frac{1}{2}m\left(arcDCB\right)$

Similarly, we know $LaTeX: \angle DCB=\frac{1}{2}m\left(arcDAB\right)$

Thus, we know that $LaTeX: \angle DAB+\angle DCB=\frac{1}{2}m\left(arcDAB+arcDCB\right)$ equation image indicator

We know that m(arcDAB) + m(arcDCB) = 360° since they comprise a whole circle, and there are 360 degrees in a circle.

Then $LaTeX: \angle DAB+\angle DCB=\frac{1}{2}\left(360^\circ\right)$

So, $LaTeX: \angle DAB+DCB=180^\circ$

Since there are $LaTeX: 360^\circ$ in any quadrilateral, then also $LaTeX: \angle ABC+\angle ADC=180^\circ$

Let's try some practice...

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