If △XYZ∼△AYB\triangle XYZ\sim\triangle AYB, and XY¯¯¯¯¯¯¯¯=24\overline{XY}=24 yards, YZ¯¯¯¯¯¯¯=30\overline{YZ}=30 yards, and YA¯¯¯¯¯¯¯¯=8\overline{YA}=8 yards, then what is the length of YB¯¯¯¯¯¯¯¯\overline{YB}?
Since triangles XYZ and AYB are similar, their corresponding sides are in proportion.
Therefore, we have:
$\frac{XY}{YA}=\frac{YZ}{YB}$
Plugging in the given values, we get:
$\frac{24}{8}=\frac{30}{YB}$
Simplifying, we get:
3 = $\frac{30}{YB}$
Cross-multiplying, we get:
$3YB = 30$
Dividing both sides by 3, we find:
YB = $\boxed{10}$ yards
PROOF: Because corresponding sides of similar triangles are proportional, ________________. Multiply both sides of the equation by segment
BC
to show that
DF=
1
2
BC
. Because corresponding angles of similar triangles are congruent,
∠ADF≅∠ABC
. If corresponding angles are congruent, then lines are parallel, so
DF
¯
¯
¯
¯
¯
¯
¯
¯
∥
BC
¯
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¯
¯
¯
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¯
.
Because corresponding sides of similar triangles are proportional, we have:
$\frac{DF}{BC}=\frac{AD}{AB}$
We can multiply both sides of the equation by BC to eliminate the fraction, resulting in:
$DF = \frac{BC \cdot AD}{AB}$
From the given information, we know that $DF=\frac{1}{2}BC$. Let's substitute that into the equation above:
$\frac{1}{2}BC = \frac{BC \cdot AD}{AB}$
We can then multiply both sides of the equation by AB to eliminate the denominators:
$\frac{1}{2}BC \cdot AB = BC \cdot AD$
Simplifying, we have:
$\frac{1}{2}AB \cdot BC = BC \cdot AD$
Dividing both sides of the equation by BC, we get:
$\frac{1}{2}AB = AD$
From this equation, we can conclude that AD is half the length of AB.
Now let's consider the angles. We have:
$\angle ADF \cong \angle ABC$
If corresponding angles of two triangles are congruent, then the lines containing those angles are parallel. So, we can conclude that DF is parallel to BC.
In summary, because corresponding sides of similar triangles are proportional, we have DF = (1/2)BC. Because corresponding angles of similar triangles are congruent, we have $\angle ADF \cong \angle ABC$, which implies that DF is parallel to BC.