Question 1066785
{{{a^2+c^2=8^2+7^2=64+49=113<121=11^2=b^2}}} ,
so ABC is an obtuse triangle,
with the obtuse angle opposite the longest side, {{{b=11}}} ,
at vertex {{{B}}} .
Triangle {{{ABC}}} , with its height {{{h=PC}}} , looks like this:
{{{drawing(300,300,-1,9,-2,8,
green(triangle(8.5,0,8,0,8.5,6.982)),
green(rectangle(8.3,0,8.5,0.2)),
red(triangle(0,0,8,0,8.5,6.982)),
arrow(8,0,10,0),arrow(0,0,-2,0),
locate(-0.2,0,red(A)),locate(8.5,0,P),
locate(7.86,0,red(B)),locate(8.4,7.45,red(C)),
locate(8.6,3.8,green(h)),locate(7.2,3.5,red(a=7)),
locate(4.25,3.5,red(b=11)),locate(3.5,0.5,red(c=8))
)}}} The projection of segment {{{AC}}} on line {{{AB}}} is segment {{{AP}}} .
Let's call the length of that projection {{{x=AP}}} , and figure out how ti find {{{x}}} .
Apply the Pythagorean theorem to right triangles {{{APC}}} and {{{BPC}}} .
For {{{BPC}}} :
{{{(x-8)^2+h^2=7^2}}}<--->{{{x^2-16x+8^2+h^2=49}}}<--->{{{x^2-16x+64+h^2=49}}}<--->{{{x^2-16x+h^2=49-64}}}<--->{{{x^2-16x+h^2=-15}}} .
For {{{APC}}} :
{{{x^2+h^2=11^2}}}<--->{{{x^2+h^2=121}}} .
Subtracting {{{x^2-16x+h^2=-15}}} from {{{x^2+h^2=121}}} , we get
{{{16x=121+15}}}--->{{{16x=136}}}--->{{{x=136/16}}}--->{{{highlight(x=17/2=8.5)}}} .