$证明:∵AE=BF$ $∴AE+EF=BF+EF$ $即AF=BE$ $∵CA⊥AB,DB⊥AB$ $∴∠A=∠B=90°$
$在Rt△ACF和Rt△BDE中$ ${{\begin{cases}{{CF=DE}}\\{AF=BE}\end{cases}}}$ $∴Rt△ACF≌Rt△BDE(HL)$ $∴∠AFC=∠BED$
$证明:∵AB⊥BC,AD⊥DC$ $∴∠B=∠D=90°$ $∴△ABC与△ACD为直角三角形$ $∵在Rt△ABC和Rt△ADC中$ $\begin{cases}{AB=AD}\\{AC=AC}\end{cases}$
$∴Rt△ABC≌Rt△ADC(HL)$ $∴∠1=∠2$
$证明:∵AE⊥EC,AF⊥BF$ $∴△AEC是直角三角形$ $△AFB是直角三角形$ $∵ 在Rt△AEC和Rt△AFB中$ $\begin{cases}{AC=AB}\\{AE=AF}\end{cases}$ $∴Rt△AEC≌Rt△AFB(HL)$
$∴∠EAC=∠FAB$ $∴∠EAC-∠BAC$ $=∠FAB-∠BAC$ $即∠1=∠2$
|
|