+0

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+5

In triangle ABC, points  D and F are on $$\overline{AB},$$ and E is on $$\overline{AC}$$  such that $$\overline{DE}\parallel \overline{BC}$$ and $$\overline{EF}\parallel \overline{CD}$$. If  AF = 7 and DF = 2, then what is BD?

second question:

In quadrilateral BCED, sides $$\overline{BD}$$ and $$\overline{CE}$$ are extended past B and C, respectively, to meet at point A. If BD = 18, BC = 18, CE = 2, AC = 7 and AB = 3, then what is DE?

Aug 3, 2024

#1
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For the first problem:

To address the problem, we will use the properties of similar triangles since lines $$DE$$ and $$EF$$ are parallel to sides $$BC$$ and $$CD$$ of triangle $$ABC$$.

Since $$DE$$ is parallel to $$BC$$ and $$EF$$ is parallel to $$CD$$, triangle $$ADF$$ is similar to triangle $$ABC$$.
The sides of similar triangles are proportional, so we can set up the following proportions:

$\frac{AD}{AB} = \frac{AF}{AC} = \frac{DF}{BC}$

We know that $$AF = 7$$ and $$DF = 2$$. Let $$BD = x$$. The entire length $$AB$$ can be expressed as:

$AB = AD + DF = AD + 2$

Let’s designate $$AD = x$$. Hence, we have $$AB = x + 2$$.

Considering the structure of triangle $$ADF$$ in relation to triangle $$ABC$$, we also need to calculate $$AC$$ in terms of $$AE$$:

Since $$EF$$ is parallel to $$CD$$, triangles $$AEF$$ and $$ACD$$ are also similar, resulting in the relationship:

$\frac{AE}{AC} = \frac{AF}{AB} = \frac{DF}{DC}$

But for our immediate task of determining the length $$BD$$ just using $$AF$$ and $$DF$$, we will not utilize this second proportion at the moment.

As both triangles $$ADF$$ and $$ABC$$ are similar due to the parallel lines, we can set the ratios of the line segments as follows:

$\frac{AD}{AB} = \frac{AF}{AC}$

But we know $$AB = AD + DF$$, substituting this gives us:

$\frac{x}{x + 2} = \frac{7}{7 + 2} = \frac{7}{9}$

Cross-multiplying yields:

$9x = 7(x + 2)$

Expanding the right-hand side:

$9x = 7x + 14$

Subtract $$7x$$ from both sides:

$2x = 14$

Solving for $$x$$:

$x = \frac{14}{2} = 7$

Thus, we find that $$AD = 7$$.

Since we established $$BD = DF = 2$$, thus:

$BD = 7 - 2 = 5$

So the length of $$BD$$ is:

$\boxed{5}$

Aug 3, 2024
#3
+760
0

Analyzing the Problem

We have a quadrilateral BCED with extended sides meeting at point A. We're given several side lengths and need to find DE.

Solution Approach

To find DE, we'll need to break down the quadrilateral into triangles and use similar triangles to establish proportions.

Breaking Down the Quadrilateral

Let's divide quadrilateral BCED into two triangles: Triangle ABC and Triangle ADE.

Using Similar Triangles

We can see that triangle ABC and triangle ADE share angle A. Additionally, angle ABC is congruent to angle ADE (alternate interior angles since BD is parallel to AE). Therefore, triangles ABC and ADE are similar.

Setting up Proportions

Since the triangles are similar, their corresponding sides are proportional. We can set up the following proportion:

AB / BD = AC / DE

Substituting Values and Solving

We know AB = 3, BD = 18, and AC = 7. Substituting these values into the proportion, we get:

3 / 18 = 7 / DE

Cross-multiplying, we get:

3 * DE = 18 * 7

DE = (18 * 7) / 3

DE = 42

Therefore, DE = 42.

Aug 4, 2024