EK⊥BC=KEK⊥BC=K

a) Ta có;

⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩ˆBHD=ˆCKE=900BD=CEˆDBH=ˆECK(=ˆACB)⇒ΔBHD=ΔCKE(ch−gn)⇒BH=CK{BHD^=CKE^=900BD=CEDBH^=ECK^(=ACB^)⇒ΔBHD=ΔCKE(ch−gn)⇒BH=CK

b) Ta có:

ΔBHD=ΔCKE(ch−gn)⇒DH=EKΔBHD=ΔCKE(ch−gn)⇒DH=EK

Lại có:

⎧⎪ ⎪⎨⎪ ⎪⎩ˆDHM=ˆEKM=900DH=EKˆHDM=ˆKEM(HD//EK(⊥BC))⇒ΔDHM=ΔEKM(g.c.g)⇒{HM=KMDM=EM⇒{HK=2HMDE=2DM⇒{HC+CK=2HMDE=2DM⇒{HC+BH=2HMDE=2DM⇒{BC=2HMDE=2DM(1){DHM^=EKM^=900DH=EKHDM^=KEM^(HD//EK(⊥BC))⇒ΔDHM=ΔEKM(g.c.g)⇒{HM=KMDM=EM⇒{HK=2HMDE=2DM⇒{HC+CK=2HMDE=2DM⇒{HC+BH=2HMDE=2DM⇒{BC=2HMDE=2DM(1)

Mặt khác:

ΔDHM;ˆDHM=900⇒DM>HM(2)ΔDHM;DHM^=900⇒DM>HM(2)

Từ (1),(2)⇒DE>BC