a) Áp dụng hệ thức lượng trong tam giác vuông ta được:
$AE.AB = AH^2$
$AF.AC = AH^2$
$\Rightarrow AE.AB = AF.AC$
b) Ta có:
$\dfrac{AC}{\sin B} = \dfrac{AC}{\dfrac{AC}{BC}} = BC$
$\dfrac{AB}{\sin C} = \dfrac{AB}{\dfrac{AB}{BC}} = BC$
Do đó:
$\dfrac{AC}{\sin B} = \dfrac{AB}{\sin C}$
c) Ta có:
$AB.AC = BC.AH = 2S_{ABC}$
$\Rightarrow BC = \dfrac{AB.AC}{AH}$
Ta được:
$BC.BE.CF$
$= \dfrac{AB.AC}{AH}.BE.CF$
$= \dfrac{(AB.BE).(AC.CF)}{AH}$
$= \dfrac{BH^2.CH^2}{AH}$
$= \dfrac{AH^4}{AH} = AH^3$
d) Ta có:
$M$ là trung điểm cạnh huyền $BC$
$\Rightarrow MA = MB = MC = \dfrac{1}{2}BC$
Theo đề ta có:
$(\sin\widehat{ACB} + \cos\widehat{ACB})^2$
$= \sin^2\widehat{ACB} + \cos^2\widehat{ACB} + 2\sin\widehat{ACB}.\cos\widehat{ACB}$
$= 1 + 2\sin\widehat{ACB}.\cos\widehat{ACB}$
$= 1 + 2.\dfrac{AH}{AC}\cdot\dfrac{HC}{AC}$
$= 1 + 2.\dfrac{AH.HC}{AC^2}$
$=1 + 2\dfrac{AH.HC}{HC.BC}$
$=1 +\dfrac{2AH}{BC}$
$= 1 +\dfrac{2AH}{2AM}$
$= 1 + \dfrac{AH}{AM}$
$= 1 + \sin\widehat{AMB}$
Do đó:
$(\sin\widehat{ACB} + \cos\widehat{ACB})^2 = 1 + \sin\widehat{AMB}$
Hay $(\sin\alpha + \cos\alpha)^2 = 1 + \sin\beta$
