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※ 引述《t0127754 (T0127754)》之銘言：
: 在ΔABC中，BC=17, CA=18, AB=19, 過ΔABC內部一點 P 向三邊分別作高D, E, F，
: 使PD⊥BC, PE⊥AC, PF⊥AB，且BD+CE+AF=27, 求BD+BF為何？
設BD=x,CE=y,AF=z
三角形PAB中
PA^2-z^2=PB^2-(18-z)^2
整理得
PA^2-PB^2=z^2-(19-z)^2...(1)
=19*(2z-19)
同理
PB^2-PC^2=x^2-(17-x)^2=17(2x-17)...(2)
PC^2-PA^2=y^2-(18-y)^2=18(2y-18)=36(y-9)...(3)
(1)+(2)+(3):
2(17x+18y+19z)=(17^2+18^2+19^2)...(4)
x+y+z=27...(5)
(5)*(6^2)-(4)
2(x-z)=6^2*3^3-(17^2+18^2+19^2)
=(2^2)*(3^5)-[(18-1)^2+18^2+(18+1)^2]
=4*243-[2*(1^2)+3*18^2]
=972-(2+324*3)=-2,x-z=-1.
原式=x-z+19=18...ans
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