如图4-169,在三角形ABC中,已知角ACB=90度,BM=MC,CP垂直于AM于点p,交AB于点D.试说明角ABM=角BPM
问题描述:
如图4-169,在三角形ABC中,已知角ACB=90度,BM=MC,CP垂直于AM于点p,交AB于点D.试说明角ABM=角BPM
答
由△ACM∽△CPM,得AM/CM=CM/PM又∵BM=CM,∴AM/BM=BM/PM又∠AMB=∠BMP(公共角),∴△ABM∽△BPM∴∠ABM=∠BPM
答
由△ACM∽△CPM,得AM/CM=CM/PM
又∵BM=CM,∴AM/BM=BM/PM
又∠AMB=∠BMP(公共角),∴△ABM∽△BPM
∴∠ABM=∠BPM