ratio of the length of right triangle are identical proof

Question

If a right triangle has an angle theta, the length of the opposite x divided by the length of adjacent y is always equal when x and y is real positive number, and it is relative to angle theta. how to prove this proposition.

Answer 1

Let in $\Delta ABC$ we have $\measuredangle ACB=90^{\circ}$ and $\measuredangle A=\theta.$

Let $D$ and $E$ be placed on the rays $AC$ and $AB$ respectively such that $\measuredangle ADE=90^{\circ}.$

Thus, $CB||DE$ and we obtain $\Delta ACB\sim\Delta ADE,$ which gives $$\frac{x}{y}=\frac{CB}{AC}=\frac{DE}{AD},$$ which says that $\frac{x}{y}$ depend on $\measuredangle A$ only.

Answer 2

You should have one or two theorems stating that if two triangles $\triangle ABC$ and $\triangle DEF$ are similar so that $\angle A \cong D$ and $\angle B \cong E$ and $\angle C \cong F$ then

1) and the proportion of any two corresponding sides are equal to the porportion of any other two corresponding sides. That is:

$\frac {AB}{DE} = \frac {BC}{EF} = \frac {AC}{DF}$.

and 2)

any the proportions of any two sides of one triangle will be the same as the proportions of the corresponding sides of the other triangle. That is:

$\frac {AB}{BC} = \frac {DE}{EF}$ and $\frac {AB}{AC} = \frac {DE}{DF}$ and $\frac {BC}{AC} = \frac {EF}{DF}$ etc.

We can prove number 2) from number 1) via the following.

Let $\triangle ABC \sim \triangle DEF$

Then $\frac {AB}{DE} = \frac {BC}{EF}$ so

$AB =\frac {BC\cdot DE}{EF}$ and

$\frac {AB}{BC} = \frac {DE}{EF}$.

We can do this for all sides.

.....

And we know if we have two triangles are such that $\triangle ABC$ and $\triangle DEF$ are such that $m\angle A = m\angle D=\theta$ and $m\angle B = m\angle E =90^\circ$ then we know $m\angle C = m\angle F = 90 -\theta$.

So the triangles are similar and the theorem 2) applies.

If the opposite angle of $BC = x$ and $AB = y$ and $EF = x'$ and $DE = y'$, then we have $\frac x{x'} = \frac y{y'}$ and by simple algebra $\frac xy = \frac {x'}{y'}$.