IMO 2014 problem 3
Dodao/la:
arhiva21. rujna 2014. Convex quadrilateral
![ABCD](/media/m/9/c/e/9ce25711ba18d9663b73c3580de4bf5a.png)
has
![\angle ABC = \angle CDA = 90^\circ](/media/m/6/a/7/6a7e34c8e7e6d5e50791c94114fd7462.png)
. Point
![H](/media/m/4/c/0/4c0872a89da410a25f00b86366efece7.png)
is the foot of the perpendicular from
![A](/media/m/5/a/e/5ae81275ee67d638485e903bdc0e9cde.png)
to
![BD](/media/m/1/1/f/11f65a804e5c922ee28a53b1df04d138.png)
. Points
![S](/media/m/c/6/3/c63593c3ec0773fa38c2659e08119a75.png)
to
![T](/media/m/0/1/6/016d42c58f7f5f06bdf8af6b85141914.png)
lie on sides
![AB](/media/m/5/2/9/5298bd9e7bc202ac21c423e51da3758e.png)
and
![AD](/media/m/6/9/6/69672822808d046d0e94ab2fa7f2dc80.png)
, respectively, such that
![H](/media/m/4/c/0/4c0872a89da410a25f00b86366efece7.png)
lies inside triangle
![SCT](/media/m/2/3/5/235e00c5222b194b1e02f896decba389.png)
and
![\angle CHS - \angle CSB = 90^\circ, \quad
\angle THC - \angle DTC = 90^\circ \text{.}](/media/m/2/f/d/2fdd56a05ab0e233fe3d13b34f6f7751.png)
Prove that line
![BD](/media/m/1/1/f/11f65a804e5c922ee28a53b1df04d138.png)
is tangent to the circumcircle of triangle
![TSH](/media/m/9/1/2/9120908f289fb15e602f1e523391ceaa.png)
.
%V0
Convex quadrilateral $ABCD$ has $\angle ABC = \angle CDA = 90^\circ$. Point $H$ is the foot of the perpendicular from $A$ to $BD$. Points $S$ to $T$ lie on sides $AB$ and $AD$, respectively, such that $H$ lies inside triangle $SCT$ and $$
\angle CHS - \angle CSB = 90^\circ, \quad
\angle THC - \angle DTC = 90^\circ \text{.}
$$
Prove that line $BD$ is tangent to the circumcircle of triangle $TSH$.
Izvor: International Mathematical Olympiad 2014, day 1