Article
MSC:
14H50
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Summary:
In the first part of the article the proof of the following theorem is given: Let point $S$ be the middle of $AB$ in the triangle $ABC$, point $O$ the intersection of $AB$ and the axis of angle $ACB$, point $P$ the foot of the perpendicular from $C$ on $AB$. If angles $ACS, SCO, OCP, PCB$ are equal, then the angle $BCA$ is the right one. In the second part, the area of right angle triangle using only the length of the axis of the right angle and of the median is derived.
References:
[1] Kuřina, F.: QED (latinsky) $\neq$ QED (anglicky). (2014). Učitel matematiky, 22(4), 206-217.
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