# Angle of Intersecting Secants

This is the idea (a,b and c are angles):

And here it is with some actual values:

In words: *the angle made by two secants (a line that cuts a circle at two points) that intersect outside the circle is half of the furthest arc minus the nearest arc.*

Why not try drawing one yourself, measure it using a protractor,

and see what you get?

It also works when either line is a tangent (a line that just touches a circle at one point). Here we see the "both are tangents" case:

That's it! You know it now.

### But How Come?

Is this magic?

Well, we can prove it if you want:

AC and BD are two secants that intersect at the point P outside the circle. What is the relationship between the angle CPD and the arcs AB and CD?

We start by saying that the angle subtended by arc CD at O is **2θ** and the arc subtended by arc AB at O is **2Φ**

By the Angle at the Center Theorem:

∠DAC = ∠DBC = θ and ∠ADB = ∠ACB = Φ

And PAC is 180°, so:

∠DAP = 180° − θ

Now use angles of a triangle add to 180° in triangle APD:

∠CPD = 180° − (∠DAP + ∠ADP)

∠CPD = 180° − (180° − θ + Φ) = θ − Φ

∠CPD = θ − Φ

∠CPD = ½(2θ − 2Φ)

Done!