Question 1159001
This is our first triangle: triangle ABC {{{drawing(400,300,-0.4,4.4,-0.3,3.3,
triangle(0,0,0,3,4,0),rectangle(0,0,0.1,0.1),
locate(4.05,0.1,A),locate(-0.15,0.1,C),
locate(-0.05,3.2,B),arc(4,0,2,2,-180,-143.1),
locate(3.1,0.35,beta),arc(0,3,2,2,36.9,90),
arc(0,3,1.8,1.8,36.9,90),locate(0.34,2.5,alpha),
locate(2,1.7,c),locate(-0.15,1.5,a),locate(1.8,-0.05,b)
)}}}
When I draw the perpendicular from C to hypotenuse AB, we will have point F, and the first triangle will be split into 2 triangles.
I will labeled them as triangle #2 and triangle #3.
{{{drawing(400,300,-0.4,4.4,-0.3,3.3,
green(line(1.38,1.84,1.46,1.78)),
green(line(0,0,1.44,1.92)),green(line(1.52,1.86,1.46,1.78)),
triangle(0,0,0,3,4,0),rectangle(0,0,0.1,0.1),
locate(4.05,0.1,A),locate(-0.15,0.1,C),
locate(-0.05,3.2,B),arc(4,0,2,2,-180,-143.1),
locate(3.1,0.35,beta),arc(0,3,2,2,36.9,90),
arc(0,3,1.8,1.8,36.9,90),locate(0.34,2.5,alpha),
locate(0.3,1.5,"#"),locate(0.4,1.5,2),
locate(0.1,1.7,triangle),locate(1.1,0.7,triangle),
locate(1.3,0.5,"#"),locate(1.4,0.5,3),
arc(0,0,1.8,1.8,-53.1,0),arc(0,0,1.6,1.6,-53.1,0),
arc(0,0,2,2,-90,-53.1),locate(0.2,0.9,beta),
locate(0.5,0.4,alpha),locate(2.15,1.9,c),
arrow(2.15,1.7,0.15,3.2),arrow(2.15,1.7,4.15,0.2),
locate(-0.15,1.5,a),locate(1.8,-0.05,b)
)}}}
Triangle ABC, triangle #2, and triangle #3 are similar triangles. They have the same shape, the same angle measures, and the same ratios of corresponding sides.
For the ratio of side opposite {{{alpha}}} to hypotenuse we have:
{{{b/c=FC/a=FA/b}}} --> {{{system(highlight(FC=ab/c),highlight(FA=b^2/c))}}}
For the ratio of side opposite {{{beta}}} to hypotenuse we have:
{{{a/c=FB/a=FC/b}}} --> {{{system(highlight(FB=a^2/c),FC=ab/c)}}}
Of course, we know that {{{FA+FB=BC=c}}} ,
We can verify that the expressions we found for {{{FA}}} and {{{FB}}} are correct, by substituting, and finding that it agrees with what we know:
From {{{c=FA+FB}}} --> {{{c=b^2/c+a^2/c}}} --> {{{c=(b^2+a^2)/c}}} --> {{{c^2=b^2+a^2}}} .
However, along the way, we find that we proved the Pythagorean theorem from our knowledge about similar triangles.
We can also prove that {{{FC^2=FA*FC}}} <--> {{{FC=sqrt(FA*FC)}}} .