Is there geometric proof of this exact problem:
How to prove geometrically that $\frac{|CD|}{|BC|} + \frac{|DE|}{|BC|} = \frac{|EC|}{|BC|}?$ Angle $BCE$ is right angle.
I don't want arithmetical proof (using $\frac{a}{c} + \frac{b}{c} = \frac{a + b}{c}$), but geometrical, using for example simillar triangles.
Much simpler version has been offered in the comment. Here is another artificial one. Since 1 has been used for labelling uploaded diagram, I let o to mean 1.
Label the regions of areas as shown.
$[\triangle BCD] + [\triangle B’DE]$
= ([0]) + ([4] + [3])
= ([0]) + ([2] + [3])
= ([0] + [2} + [3])
= $[\triangle BCE]$
Then, $0.5BC \times CD + 0.5BC \times DE = 0.5BC \times CE$
Result follows after dividing throughout by $0.5(BC)^2$.