Question

如图,已知P是正方形的边bc上的点,且bp=3pc,q是bc的中点,求证;角baq=½∠dap

Answer 1

<DAP = <APB
tan BAQ = 1/2
tan <DAP = 4/3
tan 2<BAQ = (1/2+1/2)/(1-1/4) = 4/3
<DAP=2<BAQ

追问

能不能不用tan来求

追答

在AB上取AM=5/8 AB，BM=3/8 AB，BQ=1/2 AB， 勾股定理得MQ=5/8 AB= AM
AMQ为等腰三角形，<BAQ=<MQA, <BMQ=2<BAQ
BQ/BM = 4/3 = AB/BP, BPA-BMQ相似，<BMQ=<BPA=<DAP = 2<BAQ