Question

如图,在△ABC中,AB=14,AC=24,AD平分∠BAC,过B作BD⊥AD,E为BC中点,则DE=

Answer 1

延长BD交AC于F,

∵AD平分∠BAC,∴∠DAB=∠DAF,

∵AD⊥BF,∴∠ADB=∠ADF=90°,

又AD=AD,

∴ΔADB≌ΔADF,

∴BD=BF,AB=AF,

∴CF=AC-AB=10,

又E为AB中点,

∴DE是ΔBCF的中位线,

∴DE=1/2CF=5.