Cho ∆ABC 3 góc nhọn; 3 đường cao AD, BE, CF cắt nhau tại H. Chứng minh:
a) ∆AEB ~ ∆AFC
b) ∆ADB ~ ∆AFH
c) ∆AEF ~ ∆ABC
d) ∆BDH ~ ∆AEH
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
a) Tính các góc của tam giác
Ta có: $BC^2 = 14^2 = 196$
$AB^2 + AC^2 = 4,5^2 + 13^2 = 20,25 + 169 = 189,25$
Vì: $BC^2 > AB^2 + AC^2$ nên $\widehat{A}$ là góc tù.
Áp dụng định lý cosin:
$\cos A = \dfrac{AB^2 + AC^2 - BC^2}{2 \cdot AB \cdot AC}$
$= \dfrac{20,25 + 169 - 196}{2 \cdot 4,5 \cdot 13}$
$= \dfrac{-6,75}{117}$
$\approx -0,0577$
Suy ra: $\widehat{A} \approx 93^\circ$
Áp dụng định lý cosin với góc $B$:
$\cos B = \dfrac{AB^2 + BC^2 - AC^2}{2 \cdot AB \cdot BC}$
$= \dfrac{20,25 + 196 -169}{2 \cdot 4,5 \cdot 14}$
$= \dfrac{47,25}{126}$
$\approx 0,375$
=> $\widehat{B} \approx 68^\circ$
Góc còn lại: $\widehat{C} = 180^\circ - 93^\circ - 68^\circ \approx 19^\circ$
Vậy: $\widehat{A} \approx 93^\circ,\ \widehat{B} \approx 68^\circ,\ \widehat{C} \approx 19^\circ$
b) Tính diện tích tam giác $ABC$
Nửa chu vi là: $p = \dfrac{4,5 + 14 + 13}{2} = 15,75$
Diện tích tam giác: $S = \sqrt{p(p-AB)(p-BC)(p-AC)}$
$= \sqrt{15,75(15,75-4,5)(15,75-14)(15,75-13)}$
$= \sqrt{15,75 \cdot 11,25 \cdot 1,75 \cdot 2,75}$
$\approx \sqrt{852,54}$
$\approx 29,2$
Vậy diện tích tam giác $ABC$ xấp xỉ: $29,2\ \text{đơn vị diện tích}$.
Tam giác ADE và tg ABC có
\(\hept{\begin{cases}\widehat{A}chung\\\\\frac{AD}{AE}=\frac{AB}{AC}\left(\frac{AD}{AB}=\cos\widehat{A}=\frac{AE}{AC}\right)\end{cases}}\)
Suy ra ADE đồng dạng ABC
=> đpcm
Xét ΔADB vuông tại D và ΔAEC vuông tại E có
\(\widehat{A}\) chung
Do đó: ΔADB\(\sim\)ΔAEC
Suy ra: AD/AE=AB/AC
hay AD/AB=AE/AC
Xét ΔADE và ΔABC có
AD/AB=AE/AC
\(\widehat{A}\) chung
Do đó: ΔADE\(\sim\)ΔABC
Suy ra: \(\widehat{ADE}=\widehat{ABC}\)