Consider Figure 1 *ABC* with line *l* parallel to *AC* and intersecting the other two sides at *D* and *E.*

**Figure 1 **Deriving the Side‐Splitter Theorem.

You can eventually prove that Δ *ABC*∼ Δ *DBE* using the *AA Similarity Postulate.* Because the ratios of corresponding sides of similar polygons are equal, you can show that

Now use *Property 4*, the *Denominator Subtracion Property.*

But *AB–DB = AD,* and BC–BE = *CE* ( *Segment Addition Postulate*). With this replacement, you get the following proportion.

This leads to the following theorem.

*Theorem 57 (Side‐Splitter Theorem):* If a line is parallel to one side of a triangle and intersects the other two sides, it divides those sides proportionally.

**Example 1:** Use Figure 2 *x.*

**Figure 2 ***Using the Side‐Splitter Theorem.*

Because *DE* ‖ *AC* in Δ *ABC* by *Theorem 57*, you get

**Example 2:** Use Figure 3 *x.*

**Figure 3 ***Using similar triangles.*

Notice that *TU* , *x*, is *not* one of the segments on either side that *TU* intersects. This means that you *cannot* apply *Theorem 57* to this situation. So what can you do? Recall that with *TU* ‖ *QR* , you can show that Δ*QRS*∼ Δ *TUS*. Because the ratios of corresponding sides of similar triangles are equal, you get the following proportion.

Another theorem involving parts of a triangle is more complicated to prove but is presented here so you can use it to solve problems related to it.

*Theorem 58 (Angle Bisector Theorem):* If a ray bisects an angle of a triangle, then it divides the opposite side into segments that are proportional to the sides that formed the angle.

In Figure 4*BD* bisects ∠ *ABC* in Δ *ABC*. By *Theorem 58*,

**Figure 4 **Illustrating the Angle Bisector Theorem.

**Example 3:** Use Figure 5 *x.*

**Figure 5 **Using the Angle Bisector Theorem.

Because *BD* bisects ∠ *ABC* in Δ *ABC*, you can apply *Theorem 58.*