Lời giải:
Hình vẽ.
a)
Ta có:
\(\angle ABI=180^0-\angle ABD=180^0-(90^0-\angle A)=90^0+\angle A\)
\(\angle KCA=180^0-\angle ACE=180^0-(90^0-\angle A)=90^0+\angle A\)
\(\Rightarrow \angle ABI=\angle KCA\)
Xét tam giác $ABI$ và $KCA$ ta có:
\(\left\{\begin{matrix} AB=KC\\ BI=CA\\ \angle ABI=\angle KCA\end{matrix}\right.\Rightarrow \triangle ABI=\triangle KCA(c.g.c)\)
\(\Rightarrow AI=KA\)
Ta có đpcm.
b)
Theo phần a vì \(\triangle ABI=\triangle KCA\Rightarrow \left\{\begin{matrix} \angle BAI=\angle CKA\\ \angle BIA=\angle CAK\end{matrix}\right.\)
\(\Rightarrow \angle BAI+\angle CAK=\angle CKA+\angle BIA\) (1)
Có:
\(\left\{\begin{matrix} \angle BAI+\angle BIA=\angle ABD=90^0-\angle BAC\\ \angle CAK+\angle CKA=\angle ACE=90^0-\angle BAC\end{matrix}\right.\)
Cộng hai vế:
\((\angle BAI+\angle CAK)+(\angle BIA+\angle CKA)=180^0-2\angle BAC\) (2)
Từ (1);(2) suy ra:
\(2(\angle BAI+\angle CAK)=180^0-2\angle BAC\)
\(\Leftrightarrow \angle BAI+\angle CAK=90^0-\angle BAC\)
\(\Leftrightarrow \angle IAK=\angle BAI+\angle BAC+\angle CAK=90^0\)
Do đó \(AI\perp AK\)
