Question 696621
A triangle of sides AB=77 feet, BC=60 feet and CA=97 feet, does not have a right angle at B.
A triangle of sides AB=77 feet, BC=59 feet and CA=97 feet, has an angle at C that is for practical purposes a right angle (not exactly, but you would not notice the difference).
{{{drawing(300,300,-1.5,78.5,-10,70,
triangle(0,0,0,59,77,0),rectangle(0,0,3,3),
locate(35,-1,77),locate(1,30,59),locate(37,35,97),
locate(-1,-1,B),locate(76,-1,A),locate(-1,65,C)
)}}} We need to get {{{drawing(300,300,-1.5,78.5,-10,70,
triangle(0,0,0,59,77,0),rectangle(0,0,3,3),
rectangle(10.5,0,13.5,3),rectangle(22.6,0,25.6,3),rectangle(38.5,0,41.5,3),
green(line(10.5,-10,10.5,80)),green(line(22.6,-10,22.6,80)),
green(line(38.5,-10,38.5,80)),
locate(-1,-1,B),locate(76,-1,A),locate(-1,65,C),
locate(12,-1,Z),locate(24,-1,Y),locate(40,-1,X),
locate(12,54,R),locate(24,44,Q),locate(40,32,P)
)}}} with
{{{area(AXP)=area(XYQP)=area(YZRQ)=area(ZBCR)=(1/4)*area(ABC)}}}
 
AXP and ABC are similar right triangles.
PA and CA are the corresponding longest sides of each triangle;
XP and BC are the corresponding shortest sides, and]=
AX and AB are corresponding sides too.

In similar triangles, the ratio of the areas equals the ratio of the lengths of the corresponding sides, squared.
In math equation: {{{area(ABC)/area(AXP)=(AB/AX)^2}}}
So {{{area(AXP)=(1/4)*area(ABC)}}} <--> {{{area(AXP)/area(ABC)=(1/4)}}} <--> {{{area(ABC)/area(AXP)=4}}}
means {{{(AB/AX)^2=4}}} <--> {{{AB/AX=sqrt(4)=2}}} or
{{{(AX/AB)^2=1/4}}} <--> {{{AX/AB=sqrt(1/4)=1/2}}}
So, if AB=77 feet, {{{AX=(1/2)AB=(1/2)*77}}}ft --> {{{AX=38.5}}}ft
 
If you look hard enough you will see that there are 4 {{{highlight(similar)}}} right triangles with a vertex at A:
ABC, AZR, AYQ, and AXP.
How are their areas related?
ABC is split into its 4 pieces of area equal to the area of AXP.
AYQ contains 2 of those pieces, and AZR contains 3.
So {{{area(AYQ)=2*area(AXP)}}} <--> {{{area(AYQ)/area(AXP)=2}}} 
and {{{area(AZR)=3*area(AXP)}}} <--> {{{area(AZR)/area(AXP)=3}}}
Using again the fact that the ratio of areas equals the ratio of the lengths of the corresponding sides, squared,
{{{(AY/AX)^2=2}}} --> {{{AY/AX=sqrt(2)}}} --> {{{AY=sqrt(2)*AX}}} --> {{{AY=38.5sqrt(2)}}} = about {{{highlight(54.4)}}} feet
{{{(AZ/AX)^2=3}}} --> {{{AZ/AX=sqrt(3)}}} --> {{{AZ=sqrt(3)*AX}}} --> {{{AZ=38.5sqrt(3)}}} = about {{{highlight(66.7)}}} feet