$A X$ and $D Y$ are altitiudes of two triangles $△ A B C$ and $△ D E F$ . Prove that $A X : D Y = A B : D E$ .

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Solution

$Δ A B C & Δ D E F a r e s i m i l a r$
they are equangular
Draw
$\begin{matrix} A \times ⊥ B C & D Y ⊥ E F N o w, Δ A \times B & Δ D y E a r e e q u i - a n g l e s b y A A A s w i m i l a r i t y A A \times B & A D Y E a r e s i m i l a r \end{matrix}$
Their corresponding sides are proportional
$\begin{matrix} \frac{A x}{d y} = \frac{x B}{y E} = \frac{A B}{D E} \frac{A x}{D y} = \frac{A B}{D E} \end{matrix}$

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Similar questions

Δ A B C \sim D E F

A B : D E = 3 : 4

A B C - D E F, A B : D E = 3 : 4

△

△

∠

∠

△ A B C =

△ D E F

△ A B C ≅ △ D E F

Two triangles ABC and DEF are such that, AB = ED, BC = DF, AC = EF

\frac{AB}{DE} = \frac{BC}{EF} = \frac{CA}{FD} = \frac{3}{4}

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