Construction of Tangent from Point to Circle/Proof 1

Proof

 * Euclid-III-17.png

Let $A$ be the given point and let $BCD$ be the given circle.

It is required that a straight line be drawn from $A$ to $BCD$.

Let the center $E$ of $BCD$ be found.

Join $AE$ and draw the circle $AFG$ with center $E$ and radius $AE$.

From $D$ let $DF$ be drawn perpendicular to $EA$.

Join $EF$ and let $B$ be the point at which $EF$ joins the circle $BCD$.

Join $AB$.

Then $AB$ is the required tangent to $BCD$.

Proof of Construction
Since $E$ is the center of the circles $BCD$ and $AFG$, $EA = EF$ and $ED = EB$.

Therefore the two sides $AE, EB$ equal the two sides $FE, ED$, and they contain a common angle at $E$.

So by Triangle Side-Angle-Side Equality, $\triangle DEF = \triangle BEA$.

Therefore $\angle EDF = \angle EBA$.

But $\angle EDF$ is a right angle, so $\angle EBA$ is also a right angle.

But $EB$ is a radius of circle $BCD$.

It follows from the porism to Line at Right Angles to Diameter of Circle that $AB$ is tangent to $BCD$.