Cevians perpendicular to the opposite sides are concurrent.
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Let ABC be a triangle, and let D,E,F, be the feet of the perpendiculars opposite A,B,C, respectively.
Points B,C,E,F are concyclic since they all lie on the circle with diameter BC. Therefore, with signed lengths, we have
AF⋅AB=AE⋅AC.
Similarly, we have
BD⋅BC=BF⋅BA,
CE⋅CA=CD⋅CB.
Multiplying these equalities, we get
AF⋅BD⋅CE=AE⋅BF⋅CD.
AFBF⋅BDCD⋅CFAF=1
Therefore, by Ceva's theorem, the lines AD,BE,CF meet at a point.
Points B,C,E,F are concyclic since they all lie on the circle with diameter BC. Therefore, with signed lengths, we have
AF⋅AB=AE⋅AC.
Similarly, we have
BD⋅BC=BF⋅BA,
CE⋅CA=CD⋅CB.
Multiplying these equalities, we get
AF⋅BD⋅CE=AE⋅BF⋅CD.
AFBF⋅BDCD⋅CFAF=1
Therefore, by Ceva's theorem, the lines AD,BE,CF meet at a point.
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