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An inner circle touches the outer one at point P. BC is any chord of the inner circle, which when extended, cuts the outer circle at points A and D. That is, the line segment ABCD is a chord of the outer circle. Prove that $\angle APB = \angle DPC$.

I've attached a (fancy) drawing of the problem.

I extended PB and PC to cut the outer circle at E and F respectively. It feels like $\triangle PAE \sim \triangle PDF$ (it would suffice for the proof), but I've been unable to prove so. This is my progress:

  1. $\angle PEA = \angle PDA = \angle PDC$ (same chord PA subtends equal angles on same side of circumference)

  2. $\angle PAD = \angle PAB = \angle PFD$ (chord PD)

And this is where I'm stuck. :(

enter image description here

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4 Answers 4

up vote 1 down vote accepted

This is a standard question, which uses the idea of homothety.

Hint: Show that $EF$ is parallel to $AD$. This is immediate by homothety.

Hence this gives us a trapezoid that is inscribed in a circle, hence is isosceles, so $AE=FD$, and we are done.

Your original intuition on similar triangles is incorrect, which is why your proof couldn't proceed.

However, what is true is that $PBA \sim PDF$. Show this. If you can do it directly, this is another proof.

Hint: You have already shown that $\angle PAD = \angle PFD$. Now do the other angle by angle chasing.

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Hint: Note that $PE:PB=PF:PC=r_1:r_2$, hence $EF\|BC$, hence $AE=DF$, hence the claim

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Have you learned about inversion? If you invert about point P, the result is trivial. The two circles become parallel lines, your blue line becomes a circle that passes through P, and the other lines through P remain unchanged.

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I came across the property - The angle between a tangent and a chord is equal to the inscribed angle on the opposite side of the chord.

This gives for chord PA: $\angle XPA = \angle PDA$ and for chord PB: $\angle XPB = \angle PCB$.

Using these gives the required result: $\angle APB = \angle XPB - \angle XPA = \angle PCB - \angle PDA = (\angle DPC + \angle PDC) - \angle PDA = \angle DPC$

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