We will measure the altitude of F and G points from AC, to see when they are the same.
To easy things I use as notation h=BD and hf, hg, hs, etc, the altitude of points F, G, S, etc.
Also the angle BAC is 2a and ACB is 2b.
tan.2b=h/DC tan.b=hp/DC From here hp=h*tan.b/tan.2b and so:
CQ/sin.2a=AC/sin.(2a+b) AC=h*sin.(2a+2b)/(sin.2a*sin.2b) so that CQ=h*sin.(2a+2b)*sin.2a/(sin.2a*sin.2b*sin.(2b+a))
To avoid excessive complesity in the text here we are already going to suppose that the angle ABC is 90°. If the enunciate of the puzzle is right we should get hf=hg
With ABC angle = 90° then 2a+2b=90 and 2a+b=90-b.
The expression for hq is then
Then hf=1/2(hp+hq)= h*(1-(tan.b)^2)/4 + 1/4(cos.2b)^2)
Multiplying and dividing for (cos.b)^2 the first summand
and directly from here
But for hg all the formulas are simmetric, changing a with b and b with a. and using the correspective sides of the triangle.
Edited on August 3, 2016, 1:10 pm
Posted by armando
on 2016-08-03 09:58:40