Question

【【相似!】】三角形ABC与】三角形A1B1C1是锐角三角形,AD,A1D1为角平分线

Answer 1

过B点作AC的平行线,交AD的延长线于E点,∴∠CAD=∠E,易证明△EBD∽△ACD,且AB=EB,∴BE∶AC=ED∶AD=BD∶DC,设：AB∶A1B1=AD∶A1D1=AC∶A1C1=k,∴BE∶ED∶DB=B1E1∶E1D1∶D1B1=k,∴△BED ∽△B1E1D1,∴∠E =∠E1,∴∠BAC =∠B1A1C 1∴△ABC ∽△A1B1C 1﹙两边对应成比例,且夹角相等﹚.