Question

锐角三角形ABC,AD⊥BC,BE⊥AC,求证∠CFD=∠ABC

Answer 1

∵ΔABC的三条高相交于一点F,

∴延长CF交AB于G,则CG也是高.

∴∠GAC+∠GCA=90°,

∵BE⊥AC,∴∠GAC+∠ABE=90°,

∴∠ABE=∠GCA,

同理：∠DBE=∠DAC,

∴∠CFD=∠DAC+∠GCA

=∠DBE+∠ABE

=∠ABC.