a) Xét \(\Delta ADB\) và \(\Delta CDE\) có ;
\(\widehat{ADB}=\widehat{CDE};\widehat{DAB}=\widehat{DCE}\)
\(\Rightarrow\) \(\Delta ADB\) ~ \(\Delta CDE\)
\(\Rightarrow\) \(\frac{AD}{CD}=\frac{BD}{DE}\) ; \(\frac{AD}{CD}=\frac{BD}{DE}\) ; \(\widehat{ABD}=\widehat{CED}\)
b) Xét \(\Delta BDE\) và \(\Delta ADC\) có :
\(\frac{AD}{CD}=\frac{BD}{DE}\) ; \(\widehat{BDE}=\widehat{ADC}\)
\(\Rightarrow\) \(\Delta BDE\) ~ \(\Delta ADC\)
\(\Rightarrow\) \(\widehat{DBE}=\widehat{DAC}\)
mà \(\widehat{DAC}=\widehat{BAD}=\widehat{BCE}\)
=> \(\widehat{DBE}=\widehat{BCE}\Rightarrow\Delta BCE\) cân tại E
c) Xét \(\Delta ADB\) và \(\Delta ACE\) có :
\(\widehat{ABD}=\widehat{CED}\) ; \(\widehat{BAD}=\widehat{EAC}\)
\(\Rightarrow\) \(\Delta ADB\) ~ \(\Delta ACE\)
\(\Rightarrow\) \(\frac{AD}{AC}=\frac{AB}{AE}\) \(\Rightarrow\) \(AD.AE=AC.AB\) (1)
Có : \(\frac{AD}{CD}=\frac{BD}{DE}\Rightarrow\) \(AD.DE=DB.CD\) (2)
Lấy (1) trừ (2) ta được :
\(AD.AE-AD.DE=AC.AB-BD.CD\)
\(\Rightarrow\) \(AD\left(AE-DE\right)=AC.AB-BD.CD\)
\(\Rightarrow AD^2=AC.AB-BD.CD\)
\(\Rightarrow AD^2+BD.CD=AC.AB\) (đpcm)
