6. 如图,点$C$,$D均在线段AB$上,且$AD= BC$,分别过$C$,$D作FC\perp AB$,$ED\perp AB$,垂足分别为$C$,$D$,连接$AE$,$BF$,连接$EF交AB于点G$,$AE= BF$.求证：$DG= CG$.

证明：∵​F C⊥AB，​​ED⊥AB，​∴​∠ADE=∠BCF=90°​
在​Rt∆ADE​和​Rt∆BCF {中}​
$​ \begin {cases}{AE=BF}\\{AD=BC}\end {cases}​$
∴$​Rt∆ADE≌Rt∆BCF(\mathrm {HL})​$
∴​ED=F C​
在​∆EDG ​和$​∆F CG_{中}​$
$​ \begin {cases}{∠EDG=∠F CG=90°}\\{∠EG D=∠FG C}\\{ED=F C}\end {cases}​$
∴$​∆EDG≌∆F CG(\mathrm {AAS})​$
∴​DG=CG​

7. 如图,在$Rt\triangle ABC和Rt\triangle A'B'C'$中,$\angle C= \angle C'= 90^{\circ}$,$AC= A'C'$,$AD与A'D'分别为边BC$,$B'C'上的中线且AD= A'D'$.求证：$Rt\triangle ABC\cong Rt\triangle A'B'C'$.

证明：∵​AD，​​A'D'​分别为​BC，​​B'C'​边上的中线
∴$​CD=\frac {1}{2}BC，$$​​C'D'=\frac {1}{2}B'C'​$
在​Rt∆ACD​和​Rt∆A'C'D'​中
$​ \begin {cases}{AC=A'C'}\\{AD=A'D'}\end {cases}​$
∴$​Rt∆ACD≌Rt∆A'C'D'(\mathrm {HL})​$
∴​CD=C'D'​
∴​BC=2CD=2C'D'=B'C'​
在​Rt∆ABC​和​Rt∆A'B'C'​中
$​ \begin {cases}{AC=A'C'}\\{∠C=∠C'=90°}\\{BC=B'C'}\end {cases}​$
∴​Rt∆ABC≌Rt∆A'B'C'(S AS)​