a) Ta có: ^DAC + ^BAD = 900; ^BAD + ^EAB = 900 => ^DAC=^EAB
^ACD + ^ABC = 900; ^ABE + ^ABC = 900 => ^ACD=^ABE
Xét \(\Delta\)ADC và \(\Delta\)AEB: ^DAC=^EAB; ^ACD=^ABE => \(\Delta\)ADC ~ \(\Delta\)AEB (g.g)
=> \(\frac{AC}{AB}=\frac{DC}{BE}\)=> \(BE.AC=AB.CD\)(đpcm).
b) Chứng minh tương tự câu a: \(\Delta\)ADB ~ \(\Delta\)AFC (g.g) => \(\frac{BD}{CF}=\frac{AB}{AC}\)
Lại có: \(\frac{BE}{CD}=\frac{AB}{AC}\) \(\Rightarrow\frac{BD}{CF}=\frac{BE}{CD}\)
Xét \(\Delta\)EBD và \(\Delta\)DCF: \(\frac{BD}{CF}=\frac{BE}{CD};\)^EBD=^DCF=900 => \(\Delta\)EBD ~ \(\Delta\)DCF (c.g.c)
=> ^BED=^CDF. Mà ^BED + ^BDE = 900 => ^CDF+^BDE=900 => ^EDF=900
=> \(\Delta\)DAF ~ \(\Delta\)EAD => \(\frac{AD}{AE}=\frac{DF}{DE}\Rightarrow\frac{AD}{DF}=\frac{AE}{DE}\Rightarrow\frac{AD^2}{DF^2}=\frac{AE^2}{DE^2}\)
\(\Leftrightarrow\frac{AD^2}{DF^2}+\frac{AD^2}{DE^2}=\frac{AE^2}{DE^2}+\frac{AD^2}{DE^2}=\frac{AE^2+AD^2}{DE^2}=\frac{DE^2}{DE^2}=1\)
\(\Leftrightarrow\frac{1}{DF^2}+\frac{1}{DE^2}=\frac{1}{AD^2}\)(đpcm).
