Đáp án + Giải thích các bước giải:

Ta có: $\widehat{BAH} = \widehat{CAM}$

$\Rightarrow \widehat{BAH} + \widehat{HAM} = \widehat{CAM} + \widehat{HAM}$

$\Rightarrow \widehat{BAM} = \widehat{CAH}$

Xét $\triangle ABH$ vuông tại $H$, có $\sin \widehat{BAH} = \dfrac{HB}{AB}$

$\Rightarrow HB = AB \sin \widehat{BAH} = AB \sin \widehat{CAM}$

Xét $\triangle ACH$ vuông tại $H$, có $\sin \widehat{CAH} = \dfrac{HC}{AC}$

$\Rightarrow HC = AC \sin \widehat{CAH} = AC \sin \widehat{BAM}$

$\Rightarrow \dfrac{HB}{HC} = \dfrac{AB \sin \widehat{CAM}}{AC \sin \widehat{BAM}}$

Kẻ $ME \bot AB$ tại $E, MF \bot AC$ tại $F$

Ta có: $S_{\triangle AMB} = \dfrac{AH \cdot BM}{2}$

$S_{\triangle AMC} = \dfrac{AH \cdot CM}{2}$

$\Rightarrow \dfrac{S_{\triangle AMB}}{S_{\triangle AMC}} = \dfrac{AH \cdot BM}{AH \cdot CM} = \dfrac{BM}{CM} = 1 (M\text{ là trung điểm của }BC)$

Xét $\triangle MEA$ vuông tại $E$, có $\sin \widehat{EAM} = \dfrac{EM}{AM}$

$\Rightarrow EM = AM \sin \widehat{EAM} = AM \sin \widehat{BAM}$

$\Rightarrow S_{\triangle AMB} = \dfrac{ME \cdot AB}{2} = \dfrac{AB \cdot AM \sin \widehat{BAM}}{2}$

Xét $\triangle MFA$ vuông tại $F$, có $\sin \widehat{MAF} = \dfrac{FM}{AM}$

$\Rightarrow FM = AM \sin \widehat{MAF} = AM \sin \widehat{CAM}$

$\Rightarrow S_{\triangle AMC} = \dfrac{MF \cdot AC}{2} = \dfrac{AC \cdot AM \sin \widehat{CAM}}{2}$

$\Rightarrow \dfrac{S_{\triangle AMB}}{S_{\triangle AMC}} = \dfrac{AB \cdot AM \sin \widehat{BAM}}{AC \cdot AM \sin \widehat{CAM}} = \dfrac{AB \sin \widehat{BAM}}{AC \sin \widehat{CAM}} = 1$

$\Rightarrow AB \sin \widehat{BAM} = AC \sin \widehat{CAM}$

$\Rightarrow \dfrac{AB}{AC} = \dfrac{\sin \widehat{CAM}}{\sin \widehat{BAM}}$

$\Rightarrow \dfrac{HB}{HC} = \dfrac{AB}{AC} \cdot \dfrac{AB}{AC} = \dfrac{AB^2}{AC^2}$

$\textbf{b$\bigg)$}$ Ta có: $\dfrac{HB}{HC} = \dfrac{AB^2}{AC^2}$

$\Rightarrow HC = \dfrac{HB \cdot AC^2}{AB^2}$

Theo định lý Pytago, ta có:

$AH^2 = AB^2 - BH^2 = AC^2 - CH^2$

$\Rightarrow AB^2 - BH^2 = AC^2 - \dfrac{HB^2 \cdot AC^4}{AB^4}$

$\Rightarrow AB^2 - AC^2 = BH^2 - \dfrac{HB^2 \cdot AC^4}{AB^4}$

$\Rightarrow AB^2 - AC^2 = \dfrac{BH^2 (AB^4 - AC^4)}{AB^4}$

$\Rightarrow AB^2 - AC^2 = \dfrac{BH^2 (AB^2 + AC^2)(AB^2 - AC^2)}{AB^4}$

$\Rightarrow AB^2 - AC^2 - \dfrac{BH^2(AB^2 + AC^2)(AB^2 - AC^2)}{AB^4} = 0$

$\Rightarrow (AB^2 - AC^2)\bigg(1 - \dfrac{BH^2(AB^2 + AC^2)}{AB^4}\bigg) = 0$

$\Rightarrow AB^2 = AC^2$ hoặc $\dfrac{BH^2(AB^2 + AC^2)}{AB^4} = 1$

$\Rightarrow AB = AC$ hoặc $AB^2 + AC^2 = \dfrac{AB^4}{BH^2}$

Xét $AB^2 + AC^2 = \dfrac{AB^4}{BH^2}$

Ta có: $BC = HB + HC = HB + \dfrac{HB \cdot AC^2}{AB^2} = \dfrac{HB(AB^2 + AC^2)}{AB^2}$

$\Rightarrow BH = \dfrac{BC \cdot AB^2}{AB^2 + AC^2}$

$\Rightarrow \dfrac{AB^2}{BH} = \dfrac{AB^2 (AB^2 + AC^2)}{BC \cdot AB^2} = \dfrac{AB^2 + AC^2}{BC}$

$\Rightarrow \dfrac{AB^4}{BH^2} = \dfrac{(AB^2 + AC^2)^2}{BC^2}$

$\Rightarrow AB^2 + AC^2 = \dfrac{(AB^2 + AC^2)^2}{BC^2}$

$\Rightarrow \dfrac{AB^2 + AC^2}{BC^2} = 1$

$\Rightarrow AB^2 + AC^2 = BC^2$

$\Rightarrow \widehat{BAC} = 90^\circ ($định lý Pytago đảo$)$