Question

Converse of angle between tangent and secant

Answer 1

Step-by-step explanation:

Theorem

Let D be a point outside a circle ABC.

Let DA be a straight line which cuts the circle ABC at A and C.

Let DB intersect the circle at B such that DB2=AD⋅DC.

Then DB is tangent to the circle ABC.

In the words of Euclid:

If a point be taken outside a circle and from the point there fall on the circle two straight lines, if one of them cut the circle, and the other fall on it, and if further the rectangle contained by the whole of the straight line which cuts the circle and the straight line intercepted on it outside between the point and the convex circumference be equal to the square on the straight line which falls on the circle, the straight line which falls on it will touch the circle.

(The Elements: Book III: Proposition 37)

Proof

Euclid-III-37.png

Let DE be drawn tangent to the circle ABC.

Let F be the center of ABC and join FB,FD,FE.

From Radius at Right Angle to Tangent, ∠FED is a right angle.

We have that DE is tangent to the circle ABC and DA cuts it.

So from the Tangent Secant Theorem AD⋅DC=DE2.

But we also have by hypothesis that AD⋅DC=DB2.

So DE2=DB2 and so DE=DB.

Also FE=FB and so DE,EF=DB,DF and DF is common.

So from Triangle Side-Side-Side Equality ∠DEF=∠DBF.

But ∠DEF is a right angle and so also is ∠DBF.

So from Line at Right Angles to Diameter of Circle DB is tangent to the circle ABC.