如图,RT△ABC中,角BAC=90,AB=AC,D为BC中点,P为BC上一点,PE⊥AC与E,PF⊥AB于F.求证:△DEF为等腰直角三角形 A/ \ F /\ \/ \ \/ \ \ \ / \ \ \/ \ \ \ E/ \ \ / / \/ \ / / \B --------------------------------------\/------------------\/
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![如图,RT△ABC中,角BAC=90,AB=AC,D为BC中点,P为BC上一点,PE⊥AC与E,PF⊥AB于F.求证:△DEF为等腰直角三角形 A/ \ F /\ \/ \ \/ \ \ \ / \ \ \/ \ \ \ E/ \ \ / / \/ \ / / \B --------------------------------------\/------------------\/](/uploads/image/z/6695905-49-5.jpg?t=%E5%A6%82%E5%9B%BE%2CRT%E2%96%B3ABC%E4%B8%AD%2C%E8%A7%92BAC%3D90%2CAB%3DAC%2CD%E4%B8%BABC%E4%B8%AD%E7%82%B9%2CP%E4%B8%BABC%E4%B8%8A%E4%B8%80%E7%82%B9%2CPE%E2%8A%A5AC%E4%B8%8EE%2CPF%E2%8A%A5AB%E4%BA%8EF.%E6%B1%82%E8%AF%81%EF%BC%9A%E2%96%B3DEF%E4%B8%BA%E7%AD%89%E8%85%B0%E7%9B%B4%E8%A7%92%E4%B8%89%E8%A7%92%E5%BD%A2+A%2F+%5C+F+%2F%5C+%5C%2F+%5C+%5C%2F+%5C+%5C+%5C+%2F+%5C+%5C+%5C%2F+%5C+%5C+%5C+E%2F+%5C+%5C+%2F+%2F+%5C%2F+%5C+%2F+%2F+%5CB+--------------------------------------%5C%2F------------------%5C%2F)
如图,RT△ABC中,角BAC=90,AB=AC,D为BC中点,P为BC上一点,PE⊥AC与E,PF⊥AB于F.求证:△DEF为等腰直角三角形 A/ \ F /\ \/ \ \/ \ \ \ / \ \ \/ \ \ \ E/ \ \ / / \/ \ / / \B --------------------------------------\/------------------\/
如图,RT△ABC中,角BAC=90,AB=AC,D为BC中点,P为BC上一点,PE⊥AC与E,PF⊥AB于F.
求证:△DEF为等腰直角三角形
A
/ \
F /\ \
/ \ \
/ \ \ \
/ \ \ \
/ \ \ \ E
/ \ \ / / \
/ \ / / \
B --------------------------------------\/------------------\/----------------- B P C
如图,RT△ABC中,角BAC=90,AB=AC,D为BC中点,P为BC上一点,PE⊥AC与E,PF⊥AB于F.求证:△DEF为等腰直角三角形 A/ \ F /\ \/ \ \/ \ \ \ / \ \ \/ \ \ \ E/ \ \ / / \/ \ / / \B --------------------------------------\/------------------\/
证明:
∵∠ABC=90,AB=AC
∴∠B=∠C=45
∵D是BC的中点
∴AD=BD=CD (直角三角形中线特性),AD⊥BC (等腰三角形三线合一)
∴∠ADC=∠ADB=90
∴∠DAC=45,∠ADF+∠BDF=90
∴∠DAC=∠B
∵PE⊥AC,PF⊥AB
∴矩形AEPF,∠PFB=90
∴AE=PF,BF=PF
∴AE=BF
∴△ADE≌△BDF (SAS)
∴DE=DF,∠ADE=∠BDF
∴∠ADF+∠ADE=90
∴∠EDF=90
∴等腰RT△DEF
图在哪儿
嗨,是济南外国语的么