Đáp án:
a)
Ta có:
$\widehat{BAD}=\widehat{BAC}+90^0$
$\widehat{CAE}=\widehat{BAC}+90^0$
$\Rightarrow \widehat{BAD}=\widehat{CAE}$
Gọi giao điểm của $BD$ và $CE$ là $I$
Xét $\triangle ABD$ và $\triangle AEC$ có:
$AB=AE$ (gt)
$\widehat{BAD}=\widehat{CAE}$ (cmt)
$AD=AC$ (gt)
$\Rightarrow \triangle ABD=\triangle AEC$ (c.g.c)
$\Rightarrow \widehat{ADB}=\widehat{ACE}$ (hai góc tương ứng)
Xét $\triangle ABD$ có
$\widehat{BAD}+\widehat{ADB}+\widehat{ABD}=180^0$
$\Rightarrow \widehat{BAC}+\widehat{CAD}+\widehat{ADB}+\widehat{ABD}=180^0$
$\Rightarrow \widehat{BAC} +\widehat{ADB}+\widehat{ABD}=90^0$
mà $\widehat{ADB}=\widehat{ACE}$ (cmt)
$\Rightarrow \widehat{BAC}+ \widehat{ACE}+\widehat{ABD}=90^0$
Xét $\triangle ABC$ có
$\widehat{BAC}+\widehat{ABC}+\widehat{BCA}=180^0$
$\Rightarrow \widehat{BAC}+\widehat{ABD}+\widehat{IBC}+\widehat{BCI}+\widehat{ACE}=180^0$
mà $\Rightarrow \widehat{BAC}+ \widehat{ACE}+\widehat{ABD}=90^0$ (cmt)
$\Rightarrow \widehat{IBC}+\widehat{BCI}=90^0$
$\Rightarrow \widehat{BIC}=90^0$
$\Rightarrow BD\bot CE$
b)
Giả sử $AB\bot DE$
$\Rightarrow \widehat{ABD}+\widehat{BDE}=90^0$
mà $\widehat{ABD}=\widehat{AEC}$ (do $\triangle ABD=\triangle AEC$-cmt)
$\Rightarrow \widehat{AEC}+\widehat{BDE}=90^0$
mà $\widehat{BDE} +\widehat{IED}=90^0$
$\Rightarrow \widehat{AEC}=\widehat{IED}$ (vô lý)
$\Rightarrow AB\not \bot DE$
