I happen to know that when the foot F of the altitude divides the hypotenuse into two lengths [tex]d[/tex] and [tex]e[/tex], the altitude [tex]h[/tex] is the geometric mean of [tex]d[/tex] and [tex]e[/tex], that is [tex]h^2= de.[/tex]
But you probably didn't know that. We'll take our triangle labeled the usual way, right angle C, hypotenuse [tex]c = d + e[/tex]. There are three right triangles, which I'll write without the benefit of the figure. Let's say sides [tex]a=BC[/tex] and [tex]d=BF[/tex] so
[tex]d^2 + h^2 = a^2[/tex]
[tex]e^2 + h^2 = b^2[/tex]
[tex]a^2 + b^2 = c^2 = (d + e)^2 = d^2 + 2de + e^2[/tex]
Adding all three equations lots of stuff cancels,
[tex]2h^2 = 2de[/tex]
[tex]h^2 = de[/tex]
That proves the geometric mean thing. In our problem we have [tex]h=8, d:e=1:4[/tex] which just means [tex]e=4d[/tex].
[tex] 8^2 = d(4d) = 4d^2[/tex]
[tex]d=4[/tex]
[tex]e=16[/tex]
[tex]b^2=e^2+h^2= 16^2+8^2=5(8^2)[/tex]
[tex]b=8\sqrt{5}[/tex]
Part a) [tex]d=4,e=16[/tex]
Part b) [tex]b=8\sqrt 5[/tex]
Check:
[tex] a^2=d^2+h^2=16+64=80=16\cdot 5[/tex]
The triangle is [tex]a=4\sqrt{5}, b=8\sqrt{5}[/tex] so [tex]c=\sqrt{16\cdot 5 + 64\cdot 5}=\sqrt{400}=20 \quad\checkmark[/tex]
