1) Có \(\widehat{ABC}=\widehat{ADC}\)
\(\Rightarrow180^0-\widehat{ABC}=180^0-\widehat{ADC}\) \(\Leftrightarrow\widehat{EBC}=\widehat{CDF}\)
Xét \(\Delta BCE\) và \(\Delta DCF\) có:
\(\Leftrightarrow\widehat{EBC}=\widehat{CDF}\)
\(\widehat{E}=\widehat{F}=90^0\)
nên \(\Delta BCE\sim\Delta DCF\left(g.g\right)\) \(\Rightarrow\dfrac{CE}{CF}=\dfrac{CB}{CD}\) \(\Leftrightarrow CE.CD=CF.CB\)
Có \(\widehat{EAF}+\widehat{ECF}=360^0-\widehat{AEC}-\widehat{AFC}=360^0-90^0-90^0=180^0\)
mà \(\widehat{BAD}+\widehat{ABC}=180^0\) (hai góc so le trong do BC//AD)
\(\Rightarrow\widehat{ECF}=\widehat{ABC}\) (1)
mà \(CE.CD=CB.CF\) (cm trên)\(\Leftrightarrow CE.AB=CB.CF\) \(\Leftrightarrow\dfrac{CE}{CB}=\dfrac{CF}{AB}\) (2)
Từ (1);(2) \(\Rightarrow\Delta ABC\sim\Delta FCE\left(c.g.c\right)\)
2. Kẻ \(DK\perp AC\) tại K
Dễ chững minh được \(\Delta ADK\sim ACF\left(g.g\right)\)
\(\Rightarrow\dfrac{AD}{AC}=\dfrac{AK}{AF}\Leftrightarrow AD.AF=AC.AK\) (*)
Dễ chứng minh được \(\Delta CDK\sim\Delta ACE\left(g.g\right)\)
\(\Rightarrow\dfrac{CK}{AE}=\dfrac{CD}{AC}\Leftrightarrow CK.AC=AE.CD\) mà DC=AB
\(\Rightarrow AB.AE=CK.AC\) (3*)
Từ (*);(2*) cộng vế với vế \(\Rightarrow AB.AE+AD.AF=AC.CK+AC.AK=AC\left(CK+AK\right)\)
\(\Rightarrow AB.AE+AD.AF=AC^2\)
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