Tangents

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kaloyankolev

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We have two circles k1(O1; r1) and k2(O2; r2), O1O2 > r1 + r2. The line t is the common external tangent of them, which meets the circle k1 to/in the point A1 and the circle k2 - to/in the point A2. The segment O102 crosses k1 in B1 and k2 in B2. The lines A1B1 and A2B2 are crossed in P. We have line l:
  1. P lies on l
  2. l is perpendicular to O1O2
  3. l crosses O1O2 and t in O and Q
We have to proof that Q is the middle of A1A2.

I've proved that the triangle B1B2P is right angle triangle. I can't do anything else and I will be very grateful, if you tell me how I can continue. :)

[FONT=MathJax_Main][/FONT][FONT=MathJax_Main][/FONT]The drawing: https://ibb.co/3yBJFvW[FONT=MathJax_Main][/FONT][FONT=MathJax_Main][/FONT]
 
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kaloyankolev said:
We have two circles k1(O1; r1) and k2(O2; r2), O1O2 > r1 + r2. The line t is the common external tangent of them, which meets the circle k1 to/in the point A1 and the circle k2 - to/in the point A2. The segment O102 crosses k1 in B1 and k2 in B2. The lines A1B1 and A2B2 are crossed in P. We have line l:
  1. P lies on l
  2. l is perpendicular to O1O2
  3. l crosses O1O2 and t in O and Q
We have to proof that Q is the middle of A1A2.

I've proved that the triangle B1B2P is right angle triangle. I can't do anything else and I will be very grateful, if you tell me how I can continue. :)

The drawing: https://ibb.co/3yBJFvW
Click to expand...

Try showing that A1PQ and A2PQ are isosceles. The reasoning may be similar to what you did to show that B1B2P is a right angle.
 
Dr.Peterson said:
Try showing that A1PQ and A2PQ are isosceles. The reasoning may be similar to what you did to show that B1B2P is a right angle.
Click to expand...
I showed that:
  • A1PQ is isosceles => A1Q = PQ
  • A2PQ is isosceles => A2Q = PQ
=> A1Q = A2Q
I'm right, aren't I?
 
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