Giải thích các bước giải:

a.Xét $\Delta AHB,\Delta AHD$ có:
Chung $AH$

$\widehat{AHB}=\widehat{AHD}(=90^o)$

$HB=HD$

$\to\Delta AHB=\Delta AHD(c.g.c)$

$\to AB=AD$

b.Ta có $AB\perp AC, DE\perp AC\to AB//DE$

Xét $\Delta HAB,\Delta HDE$ có:

$HB=HD$

$\widehat{HBA}=\widehat{HDE}$ vì $AB//DE$

$AB=DE$

$\to\Delta HAB=\Delta HED(c.g.c)$

$\to HA=HE$

$\to H$ là trung điểm $AE$

c.Ta có $AH\perp BC\to CD\perp AE$

              $ED\perp AC$

$\to D$ là trực tâm $\Delta ACE$

$\to AD\perp CE$

Xét $\Delta AHC,\Delta EHC$ có:

Chung $CH$

$\widehat{AHC}=\widehat{CHE}(=90^o)$

$HA=HE$

$\to\Delta CHA=\Delta CHE(c.g.c)$

$\to\widehat{ACH}=\widehat{ECH}$

$\to\widehat{KCD}=\widehat{FCD}$

Xét $\Delta CDK,\Delta CFD$ có:

Chung $CD$

$\widehat{KCD}=\widehat{DCF}$

$CK=CF$

$\to\Delta CDK=\Delta CDF(c.g.c)$

$\to\widehat{DFC}=\widehat{DKC}=90^o$

$\to DF\perp CE(2)$

Từ $(1), (2)\to A , D, F$ thẳng hàng 

Đáp án:

Giải thích các bước giải:

a.Xét ΔAHB,ΔAHDΔAHB,ΔAHD có:
Chung AHAH

ˆAHB=ˆAHD(=90o)AHB^=AHD^(=90o)

HB=HDHB=HD

→ΔAHB=ΔAHD(c.g.c)→ΔAHB=ΔAHD(c.g.c)

→AB=AD→AB=AD

b.Ta có AB⊥AC,DE⊥AC→AB//DEAB⊥AC,DE⊥AC→AB//DE

Xét ΔHAB,ΔHDEΔHAB,ΔHDE có:

HB=HDHB=HD

ˆHBA=ˆHDEHBA^=HDE^ vì AB//DEAB//DE

AB=DEAB=DE

→ΔHAB=ΔHED(c.g.c)→ΔHAB=ΔHED(c.g.c)

→HA=HE→HA=HE

→H→H là trung điểm AEAE

c.Ta có AH⊥BC→CD⊥AEAH⊥BC→CD⊥AE

              ED⊥ACED⊥AC

→D→D là trực tâm ΔACEΔACE

→AD⊥CE→AD⊥CE

Xét ΔAHC,ΔEHCΔAHC,ΔEHC có:

Chung CHCH

ˆAHC=ˆCHE(=90o)AHC^=CHE^(=90o)

HA=HEHA=HE

→ΔCHA=ΔCHE(c.g.c)→ΔCHA=ΔCHE(c.g.c)

→ˆACH=ˆECH→ACH^=ECH^

→ˆKCD=ˆFCD→KCD^=FCD^

Xét ΔCDK,ΔCFDΔCDK,ΔCFD có:

Chung CDCD

ˆKCD=ˆDCFKCD^=DCF^

CK=CFCK=CF

→ΔCDK=ΔCDF(c.g.c)→ΔCDK=ΔCDF(c.g.c)

→ˆDFC=ˆDKC=90o→DFC^=DKC^=90o

→DF⊥CE(2)→DF⊥CE(2)

Từ (1),(2)→A,D,F(1),(2)→A,D,F thẳng hàng