Angle of Intersecting Secants

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

Circle with two secants intersecting outside, showing angles a, b, and c.

And here it is with some actual values:

Example with far arc 80 degrees, near arc 30 degrees, and intersection angle 25 degrees.

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:

Two tangents intersecting outside a circle with far arc 260 degrees and near arc 100 degrees.

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's the relationship between the angle CPD and the arcs AB and CD?

Circle with two secants intersecting at point P outside the circle, forming triangle APD.

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!

7385, 7386, 7387, 7388, 7389, 13611, 13612, 13613, 13614, 13615