Bai 1:
Xet ΔABC can tai C:
⇒∠CBA=∠CAB , CA=CB
Xet ΔCBH va ΔCAH
CA=CB
∠CBA=∠CAB
∠CHA=∠CHB=90
⇒ΔAHC=ΔBHC ( ch-gn)
Bai 2:
a, Xet ΔABH va ΔACH
AB=AC
∠AHC=∠AHB =90
AH la canh chung
⇒ΔAHB=ΔAHC ( ch-cgv)
⇒HB=HC( 2 canh tuong ung bang nhau)
b, Co ΔAHB=ΔAHC ( cau a)
⇒∠BAH=∠CAH( 2 goc tuong ung bang nhau )
c, Xet ΔAEH va ΔADH
AH la canh chung
∠BAH=∠CAH (cmt)
∠AEH=∠ADH =90
⇒ΔAEH=ΔADH( ch-gn)
⇒AE=AD( 2 canh tuong bang nhau)
⇒ΔAED can tai A
Bai 3
a, Xet ΔMDE va ΔMEF
ME la canh chung
DE=EF ( gt)
∠MED=∠MEF=90
⇒ΔMDE=ΔMFE( cgv-cgv)
b, Xet ΔDEN va ΔEFN
DE=EF (gt)
∠DEN=FEN=90
EN la canh chung
⇒ΔDEN=ΔFEN(cgv-cgv)
⇒DN=NF( 2 canh tuong ung bang nhau)
