perhatikan gambar diatas.segitiga-segitiga di dalam segitiga kongruen.panjang AD adalah

ABD kongruen dengan AED dan GED

anggap saja AD = DC = X
dan BD = DE = Y
dan
[tex]a^{2} + b^{2} = c^{2} \\ AB^{2} + BD^{2} = AD^{2} \\ 12^{2} + Y^{2} = X^{2} \\ 12^{2} = X^{2} - Y^{2} = (X+Y)(X-Y) \\ 144 = (X+Y)(X-Y) [/tex]


pada segitiga besar ABC
berlaku,



[tex] a^{2} + b^{2} = c^{2} \\ AB^{2} + BD^{2} = AD^{2} \\ 12^{2} + Y^{2} = (12+12)^{2} \\ 144 + ( [tex] ((X+Y)^{2}) = 24^{2} \\ 144 + ( X^{2} + 2XY + Y^{2}) = 576 \\ 576-144 = ((X+Y)^{2}  ) \\ 432 = ( (X+Y)^{2})[/tex]

X+Y = [tex] \sqrt{432} = 12 \sqrt{3} [/tex]
X-Y = (X+Y)(X-Y) / (X+Y) = [tex] \frac{14}{12 \sqrt{3}} x \frac{12 \sqrt{3}}{12 \sqrt{3}} = \frac{48 \sqrt{3}}{432} = \frac{\sqrt{3}}{9} [/tex]

X+Y =  [tex]12 \sqrt{3} [/tex]
X-Y = [tex]\frac{\sqrt{3}}{9} [/tex] +
------------------------------------------------
2X = [tex]12 \sqrt{3} + \frac{\sqrt{3}}{9} [/tex] 
      = [tex] \frac{108\sqrt{3}  + \sqrt{3}}{9} [/tex] 
      = [tex] \frac{109\sqrt{3}}{9} [/tex] 
X = [tex] \frac{109\sqrt{3}}{9} : 2 [/tex] =  [tex] \frac{109\sqrt{3}}{18} [/tex]