Question

如图,在三角形ABC中,AD平分角BAC,BE垂直AC于E,交AD于F,试说明角2=1\2(角ABC+角C)

Answer 1

∵AD平分∠BAC

∴∠DAC=1/2∠BAC

∵∠BAC=180°-(∠ABC+∠C)

∴∠FAE=∠DAC=1/2[180°-(∠ABC+∠C)]=90°-1/2(∠ABC-∠C)

∵BE⊥AC

∴∠BEA=90°

∴∠2(∠AFE)=90°-∠FAE

=90°-90°+1/2(∠ABC+∠C)

=1/2(∠ABC+∠C)

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