已知在△ABC与△A1B1C1中,AB=A1B1,BC=B1C1,∠BAC=∠B1A1C1=100°,求证：△ABC≌△A1B1C1

过点A做AD⊥BA,过A1做A1D1⊥B1A1则在Rt△ABD和Rt△A1B1D1中AB=A1B1 ∠BAD=∠B1A1D1 AD=A1D1 ∴Rt△ABD≌Rt△A1B1D1 SAS∵Rt△ABD≌Rt△A1B1D1 ∴∠BAD=∠B1A1D1 ∠BAC=∠B1A1C1=100° ∴∠CAD=∠C1A1D1=10°又∵∠AB...