Awesomeness of Angle Chasing and Cyclic Quadrilaterals
by shiningsunnyday, Mar 25, 2016, 2:58 AM
Angle chasing is really awesome. But of course, the basics (e.g. labelling angles, isosceles triangles, parallelograms.) are mostly limiting. So this is where cyclic quadrilaterals come in, when direct angle chasing becomes pointless. Another note is that most stuff below are pretty introductory, since I'm not good enough to do advanced problems yet, but I'll get to some hard angle-chasing/cyclic quadrilateral problems in the near future.
Another nice application of this tangent property is the first Brocard point of a triangle. I strongly recommend you guys check out the proof online!
Finally, here's a bonus question that I haven't had time to solve yet. As a bonus, whoever solves the following problem gets to propose what my next blog topic will be about! (Try to keep it within the realm of hard pre-olympiad/entry-olympiad Alg, Geo, or some Combo topic).
degrees, it's cyclic.![[asy]
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[/asy]](http://latex.artofproblemsolving.com/3/c/a/3ca3da7776560689e871c90deda1001f4acaefa6.png)
![[asy]
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[/asy]](http://latex.artofproblemsolving.com/7/8/4/7845615566b1f4a13ff295b2070c3f6ad6d30ae5.png)
,
and 
sweep out the same arc, and have the same angle. Finally, 
.![[asy]
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[/asy]](http://latex.artofproblemsolving.com/5/e/6/5e66a86af3d328881e3314406e92b181647005c8.png)
on side 
and 
on 
of a random triangle. Call the intersection of the circumcircles (that's not 
and 
to be 
.
lie on the same circle.![[asy]
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[/asy]](http://latex.artofproblemsolving.com/a/5/6/a568d7e60e4242e955545703adc4ff3e56095eeb.png)
and 
add up to 
![[asy]
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[/asy]](http://latex.artofproblemsolving.com/b/1/1/b11dbcb86b90b96ce40803809c1a4ec019dadf88.png)
an arbitrary point 
is selected. The tangent in 
meets 
.
is defined analogously. Prove that 
![[asy]
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[/asy]](http://latex.artofproblemsolving.com/8/5/b/85b8d5e2eac8c22a97884440f837ebc2a18195bc.png)
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[/asy]](http://latex.artofproblemsolving.com/4/7/7/4776110780647e87b624afb7a2d6e65612ad4595.png)
,
and 
by the property. 
by the property as well. The two red angles are also equal by the property.
(by vertical angles).![[asy]
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[/asy]](http://latex.artofproblemsolving.com/d/e/8/de88c9b442f3d705465cd748da1222b77bfc097e.png)
Aha! 
and 
because of the isosceles triangles! Our angle chasing is finally paying off. Adding 
to each side of: 
hence proving the result.Another nice application of this tangent property is the first Brocard point of a triangle. I strongly recommend you guys check out the proof online!
Finally, here's a bonus question that I haven't had time to solve yet. As a bonus, whoever solves the following problem gets to propose what my next blog topic will be about! (Try to keep it within the realm of hard pre-olympiad/entry-olympiad Alg, Geo, or some Combo topic).
lies inside triangle 
lie on the segments 
,
,
,
pass through a common point.
May 2018
April 2018