Question

证明：在RtΔABC中,∠A=90°,三角形内切圆切BC于D,则S△ABC=BD×CD

Answer 1

证明：设△ABC的内切⊙O与AC切于E,与AB切于F,连接OD,OE,OF则BD=BF,CD=CF,AE=AE（从圆外一点引圆的两条切线长相等）∵OE⊥AC,OF⊥AB,∠A=90°∴四边形AEOF是矩形∵OE=OF∴四边形AEOF是正方形设OD=OE=OF=AE=AF=r则AB=BF+AF=BD+rAC=CE+AE=CD+r根据勾股定理AB²+AC²=BC²（BD+r）²+（CD+r）²=（BD+CD）²BD²+2BDr+r²+CD²+2CDr+r²=BD²+2BD×CD+CD²BD×CD=BDr+CDr+r²∵S四边形BDOF=S△BDO+S△BFO=2S△BDO=BDr S四边形CDOE=2S△CDO=CDr S正方形AEOF=r² S△ABC=S四边形BDOF+S四边形CDOE+S正方形AEOF∴S△ABC=BD×CD