I'm solving a homework problem and I need to explicitly show that 2 angles are distinct. The problem is as follows:

Given a circle of radius $4r$ which is tangent to a circle of radius $2r$, take a line parallel to the axis which connects both circle centers that is $r$ distance apart from said axis. We call the $2$ points from the intersection of this secant line with both circles that are furthest apart to be $A$ and $B$, and we call the other pair of points $C$ and $D$. Show that the angle between $A,B$ and the tangency point is greater than the angle between $C,D$ and the tangency point.

I made the following diagram of the situation. enter image description here

It seems fairly obvious that the triangle$\triangle PAB$ contains the smaller triangle $\triangle PDC$, but I don't know how to justify this. I was thinking of explicitly calculating the equations for both circles and the secant line to calculate the coordinates of the points, but it seems like overkill to me. Does anyone know a simple argument I could use to show that $\angle APB \neq \angle CPD$?

  • $\begingroup$ A) I would have just called it an obvious fact about points $A, D, C, B$ on a line in that order. B) Note that $ \angle APB + \angle DPC = 180^\circ$, which is what I'm hoping this leads up to. $\endgroup$
    – Calvin Lin
    Oct 31 '20 at 19:10
  • 1
    $\begingroup$ ∠APB= ∠APD+∠DPC+∠BPC, since the other are positive values ∠APB > ∠DPC $\endgroup$
    – Moti
    Nov 1 '20 at 2:22

It does seem like overkill, but unless you are required to use the numbers given, which my figure reflects but I do not make use of, an explicit proof might be to draw the tangent $PK$.

Then since $K$ lies between $D$ and $C$,$$\angle APK>\angle DPK$$Likewise,$$\angle BPK>\angle CPK$$Therefore by addition$$\angle APB>\angle DPC$$ angle comparison


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