4.12: Missing Side Lengths and Similarity or the “Side-Splitting Theorem”
Learning Objectives
- Identify proportional segments when two sides of a triangle are cut by a segment parallel to the third side.
- Divide a segment into any given number of congruent parts.
Dividing Sides of Triangles Proportionally
A midsegment of a triangle:
- is parallel to one side of a triangle and
- divides the other two sides into congruent halves (because it connects the midpoints of those two sides).
A midsegment is _____________________________ to one side of a triangle and connects the _______________________________ of the other two sides.
Therefore, the midsegment divides those two “other” sides proportionally.
Remember, When two parallel lines are cut by a transversal, corresponding angles are congruent (or \begin{align*}\angle 1 \cong \angle 2\end{align*}
A midsegment may look like \begin{align*}\overline{EF}\end{align*}
Example 1
Explain the meaning of "the midsegment divides the sides of a triangle proportionally."
Suppose each half of one side of a triangle is \begin{align*}x\end{align*}
- One side of the triangle is divided in the ratio \begin{align*}x : x\end{align*}
x:x and the other side in the ratio \begin{align*}y : y\end{align*}y:y . - Both of these ratios are equivalent to 1 : 1 and to each other.
We see that a midsegment divides two sides of a triangle proportionally. But what about some other segment?
Triangle Proportionality Theorem
If a line parallel to one side of a triangle intersects the other two sides, then it divides those sides into proportional segments.
\begin{align*}\rightarrow\end{align*}
The diagram above demonstrates this theorem as well as the definition of a midsegment.
- If a line intersects two sides of a triangle and is parallel to the third side, then it splits the intersected sides into _____________________________________ segments.
- This is called the Side- ________________________ Theorem because it splits two sides of a triangle.
Check out the proof of the Side-Splitting Theorem on the following page.
Proof:
Given: \begin{align*}\Delta ABC\end{align*}
Prove: \begin{align*}\frac{AD}{DB} = \frac{CE}{EB}\end{align*}
Statement | Reason |
---|---|
1. \begin{align*}\overline{DE} \ \| \ \overline{AC}\end{align*} |
1. Given |
2. \begin{align*}\angle 1 \cong \angle 3, \angle 2 \cong \angle 4\end{align*} |
2. Corresponding angles are congruent |
3. \begin{align*}\Delta ABC : \Delta DBE\end{align*} |
3. AA Similarity Postulate |
4. \begin{align*}AD + DB = AB, CE + EB= CB\end{align*} |
4. Segment Addition Postulate |
5. \begin{align*}\frac{AB}{DB} = \frac{CB}{EB}\end{align*} |
5.Corresponding side lengths in similar triangles are proportional |
6. \begin{align*}\frac{AD + DB}{DB} = \frac{CE + EB}{EB}\end{align*} |
6. Substitution |
7. \begin{align*}\frac{AD + DB}{DB} = \frac{AD}{DB} + \frac{DB}{DB} = \frac{AD}{DB}+1\end{align*} |
7. Algebra |
8. \begin{align*}\frac{AD}{DB} + 1 = \frac{CE}{EB} + 1\end{align*} |
8. Substitution |
9. \begin{align*}\frac{AD}{DB} = \frac{CE}{EB}\end{align*} |
9. Subtraction Property of Equality |
Can you see why we wrote the proportion in step #6 this way, rather than as \begin{align*}\frac{DB}{AD + DB} = \frac{EB}{CE + EB}\end{align*}
It is because \begin{align*}\frac{x + y}{z} = \frac{x}{z} + \frac{y}{z}\end{align*}
Note: The converse of this theorem is also true.
Converse of the Triangle Proportionality Theorem
If a line divides two sides of a triangle into proportional segments, then the line is parallel to the third side of the triangle.
Example 2
In the diagram below, \begin{align*}UV : NP = 3 : 5\end{align*}
What is an expression in terms of \begin{align*}x\end{align*}
According to the Triangle Proportionality Theorem,
\begin{align*}\frac{3}{5} = \frac{MU}{MU+ 3x}\end{align*}
Use cross multiplication so
\begin{align*}3 (MU + 3x) &= 5 (MU)\\
3MU + 9x &= 5MU\\
2MU &= 9x\\
MU &= \frac{9x}{2} = 4.5x\end{align*}
Now find the length of \begin{align*}\overline{MN}\end{align*}
\begin{align*}MN &= MU + UN = 4.5x + 3x\\
MN &= 7.5x\end{align*}
Reading Check:
1. Fill in the blanks:
A midsegment is ________________________ to one side of a triangle and divides the other two sides of the triangle into ______________________________ halves.
2. True/False: A midsegment connects the midpoints of all three sides of a triangle.
3. True/False: If a line divides two sides of a triangle into proportional segments, then that line must be parallel to the third side of the triangle.
Parallel Lines and Transversals
Example 3
Look at the diagram below:
\begin{align*}k, m, n, p,\end{align*}
\begin{align*}a, b, c,\end{align*}
\begin{align*}k, m,\end{align*}
We’re given that lines \begin{align*}k, m,\end{align*}
Lines \begin{align*}k, m,\end{align*}
Lines \begin{align*}p\end{align*}
Following the Triangle Proportionality Theorem, it is also true that:
The segment lengths on one transversal are proportional to the segment lengths on the other transversal.
The transversals have _________________________________ segment lengths, so we can set up the following proportions:
\begin{align*}\frac{a}{b} = \frac{c}{d}\end{align*}
Reading Check:
In the diagram above, which lines are parallel?
which lines are transversals?
Example 4
What we discovered in Example 3 can be broadened to any number of parallel lines that cut any number of transversals. When this happens, all corresponding segments of the transversals are proportional!
The diagram below shows several parallel lines, \begin{align*}k_1, k_2, k_3,\end{align*}
The \begin{align*}k\end{align*} lines are all parallel.
Now we have lots of proportional segments.
For example: \begin{align*}\frac{a}{b} = \frac{d}{e}, \frac{a}{c} = \frac{g}{i}, \frac{b}{h} = \frac{a}{g}, \frac{c}{f} = \frac{b}{e}\end{align*}, and many more.
Reading Check:
In the space below, draw a picture (it can be a shape or a group of lines) that shows that the Triangle Proportionality Theorem (or Side-Splitting Theorem) is true. Be sure to label all the segments in your picture that you will use to make your proportion. Clearly write the proportion next to your picture.
\begin{align*}{\;}\end{align*}
\begin{align*}{\;}\end{align*}
\begin{align*}{\;}\end{align*}
\begin{align*}{\;}\end{align*}
\begin{align*}{\;}\end{align*}
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