Question 192713
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If triangle ABC has dimensions {{{a^2+b^2=c^2}}}, we want to prove angle BCA is a Right Triangle.


Remember a Square is formed by four Right Triangles, being each triangle has the same dimensions as the other triangles.



See figure below:
{{{drawing(400,400,-1,8,-1,8,line(0,0,0,6),line(0,6,6,6),line(6,6,6,0),line(6,0,0,0),green(line(0,2,2,6)),green(line(2,6,6,4)),green(line(6,4,4,0)),green(line(4,0,0,2)),green(locate(2,5.7,90^0)),green(locate(.3,2.5,90^o)),green(locate(3.5,.9,90^o)),green(locate(5,4,90^o)),line(0,.3,.3,.3),line(.3,.3,.3,0),line(5.7,0,5.7,.3),line(5.7,.3,6,.3),line(5.7,6,5.7,5.7),line(5.7,5.7,6,5.7),line(.3,6,.3,5.7),line(.3,5.7,0,5.7),locate(5,3,c),locate(1.5,1.8,c),locate(2,-.2,a),locate(5,-.2,b),locate(6.2,2,a),locate(6.2,5,b))}}}


We get the Area of the outer Square:
{{{A[o]=(a+b)^2}}}
Also,
{{{A[o]=c^2+highlight(4(1/2)(ab))}}}


*Note: {{{4(1/2)(ab)}}} ----> Area of 4 Right Triangles outside the inner square.


Therefore,
{{{A[o]=A[o]}}}
{{{(a+b)^2=c^2+(cross(4)2(1/cross(2)1)(ab))}}}
{{{a^2+2ab+b^2=c^2+2ab}}}
{{{a^2+cross(2ab)+b^2=c^2+cross(2ab)}}}
{{{red(a^2+b^2=c^2)}}}


Then, Triangle ABC has sides {{{a^2+b^2=c^2}}}


{{{drawing(200,200,-1,3,-1,5,triangle(0,0,1.5,4,1.5,0),line(1.3,0,1.3,.3),line(1.3,.3,1.5,.3),locate(.7,-.15,b),locate(1.7,2,a),locate(.7,2.9,c),red(locate(1.5,4.4,B)),red(locate(1.7,.12,C)),red(locate(-.15,0,A)))}}} ---> Angle BCA is Right Triangle


Thank you,
Jojo</font>