# No algebra please, we are geometers

Given a right triangle with sides $$ABC$$ make two more right triangles using sides $$A$$ and $$C$$ (long side) and a new long side $$x$$ (same for both new triangles). By Pythagoras the implied third sides will have lengths $$a$$ and $$c$$ such that $$a^2+A^2 = x^2 = c^2+C^2$$.

Now using some algebra I can show that if we can form a triangle with sides $$aBc$$ it must be right, too, viz.: $$B^2+c^2 = B^2 + x^2 - C^2 = x^2 - A^2 = a^2$$

But that feels just wrong like instrument flying on a bright day.

Can you

• either rearrange the figure in such a way as to make it ($$aBc$$ is right) obvious
• or make a direct geometric argument
• or a combination of both?

Note on the figure. By unfortunate coincidence (pun intended) the purple circle appears to pass through $$\angle AB$$. That is not necessarily the case. The circle is the one of radius $$c$$ around $$\angle BC$$

• In case you want to improve the diagram, here are some observations: 1: it's very hard to follow the text and find the corresponding parts in the image, there's too much clutter. Maybe start with a much simpler diagram, and then add the relevant elements to the picture as the text explanation progresses? 2: The points aren't labeled, so it's cumbersome to write about the diagram. 3: The circle with radius c doesn't pass through the single point (at the angle of x and c) that's initially known to be exactly at distance c from the centre.
– Bass
Aug 18, 2020 at 12:44
• @Bass thanks, these are valid observations. Aug 18, 2020 at 14:23

Suppose we choose a point in the plane through $$B$$ perpendicular to the plane of the triangle. This creates three new triangles. The triangle upon $$A$$ is always right. (This is $$Axa$$.) The triangle upon $$C$$ is right at $$BC$$ if and only if the the point is directly above vertex $$BC$$ (ie the line through the new point and vertex $$BC$$ is perpendicular to the plane of the original triangle). (This is $$Cxc$$.) In this case, the triangle upon $$B$$ is clearly right as well (also at $$BC$$). (This is $$aBc$$.)