Let ABC be a triangle. Let S be the circle through B tangent to CA at A and let T be the circle through C tangent to AB at A. The circles S and T intersect at A and D. Let E be the point where the line AD meets the circle ABC. Prove that D is the midpoint of AE.

  • $\begingroup$ Well, we can have no idea... might be a typo $\endgroup$ – QuIcKmAtHs Jan 29 '18 at 22:09
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    $\begingroup$ Or perhaps "the circle ABC" means the circumcircle of the triangle ABC? After all, 3 points uniquely define a circle $\endgroup$ – glowstonetrees Jan 29 '18 at 22:10
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    $\begingroup$ Three non-colinear points always define a unique circle. So the question is not wrong, perhaps a little loosely worded. $\endgroup$ – Moriarty Jan 29 '18 at 22:11
  • $\begingroup$ Well, thank you. Will consider it's circumscribed as you suggest and try to prove accordingly. Thanks again :'D $\endgroup$ – User7 Jan 29 '18 at 22:16

I hope this is correct...

Assuming a triangle $A,B,C$. it seems Circle $S$ has it's origin at the midpoint of the line $AB$ (radius $\cfrac{AB}{2}$). Similarly Circle $T$ has its origin at the midpoint of $BC$ (radius $\cfrac{BC}{2}$).

For the circles to intersect at $D$ means $AB=BC$ and $AE$ is the hypotenuse of the triangle $\Bigg(\bigg(\cfrac{AB}{2}\bigg)^{2}+\bigg(\cfrac{BC}{2}\bigg)^{2}\Bigg{)}^{1/2}$.

For $D$ to be the midpoint of $AE$. $D$ would need to be the origin of the circle $ABC$ with a radius of the hypotenuse above.

  • $\begingroup$ Oh, thanks! I just have one question, why did you say that Circle S has its origin at the midpoint of AB, and Circle T has its origin at the midpoint of BC? $\endgroup$ – User7 Feb 2 '18 at 16:04
  • $\begingroup$ Hi, Sorry, since there was no information given about the angles of the triangle. I had to work with a right triangle with AC perpendicular to AB, then.....for a radius to pass thru B and be perpendicular to the line AC it must start at the midpoint of BC. If it's not a right triangle....I myself cannot see how to proceed. So either I don t have the ability to see another approach, or it's a very poorly worded question. $\endgroup$ – Henry Feb 3 '18 at 19:33
  • $\begingroup$ Hi, thanks again! Yeah I got your point. And I, too, think that the question is poorly worded. $\endgroup$ – User7 Feb 5 '18 at 23:51

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