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All courses Math II · G-SRT.4 49 of 73
Prove triangle-similarity theorems, including proportional segments and the Pythagorean Theorem via similarity.

Use the side-splitter theorem to find a missing segment

Problem
Use the side-splitter theorem to find the missing segment: DE parallel BC, AD=4, DB=6, AE=x, EC=9
Unworked geometric configuration showing only the supplied segment, angle, parallel, or right-angle evidence. Open full size
Your answer
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Hint

Set up the side-splitter proportion using the two split segments on each side: \(AD/DB = AE/EC\).

When a line is parallel to one side of a triangle, it divides the other two sides proportionally.

Solution walkthrough

01

State why the side-splitter theorem applies

\[D~\text{on}~AB;~E~\text{on}~AC;~DE~\text{parallel}~BC\]

A segment joining two sides of a triangle and parallel to the third side divides those two sides proportionally.

02

Match upper and lower split segments

\[AD/DB=AE/EC~->~4/6=x/9\]

Both ratios compare the segment above the parallel line with the segment below it on the same triangle side.

03

Solve the proportion

\[4(9)=6x~->~36=6x~->~x=6\]

Cross products are equal; division by 6 isolates the missing length.

04

Interpret and check

\[4/6=6/9=2/3\]

The solved length gives equal split-side ratios and represents segment AE, so the missing segment is 6 length units.

+

Another way

  1. Simplify 4/6 to 2/3, then solve x/9=2/3 to get x=6.

!

Common mistake

Do not mix a split segment with a whole side. The proportion here uses upper/lower on both sides: AD/DB and AE/EC.

Completed theorem configuration with justified marks, proportion, and conclusion.
Parallel side splitter creates proportional split segments and determines the missing length.