\(\Delta ABD\) cân tại \(B\) (do \(AB=BD\)) nên \(\widehat{ADB}=\frac{180^0-\widehat{B}}{2}\).
\(\Delta AEC\) cân tại \(C\) (do \(CE=CA\)) nên \(\widehat{AEC}=\frac{180^0-\widehat{C}}{2}\).
Do đó \(\widehat{ADB}+\widehat{AEC}=\frac{360^0-\left(\widehat{B}+\widehat{C}\right)}{2}=\frac{360^0-90^0}{2}=\frac{270^0}{2}=135^0\)
\(\Rightarrow90^0+\widehat{DAE}=135^0\)
\(\Rightarrow\widehat{DAE}=135^0-90^0\)
\(\Rightarrow\widehat{DAE}=45^0\)
Vì AB=BD => tam giác ABD cân tại B
=> \(\widehat{ADB}=\dfrac{180^0-\widehat{B}}{2}\)
Vì CE = CA => tam giác AEC cân tại C
=> \(\widehat{AEC}=\dfrac{180^0-\widehat{C}}{2}\)
Do đó
\(\widehat{ADB}+\widehat{AEC}=\dfrac{360^0-\left(\widehat{B}+\widehat{C}\right)}{2}\)
= \(\dfrac{360^0-90^0}{2}\)= \(\dfrac{270^0}{2}=135^0\)
=> \(90^0+\widehat{DAE}=135^0\)
=> \(\widehat{DAE}=135^0-90^0=45^0\)
