Problem

Complete the side-splitter proof from \(\triangle~ADE~∼~\triangle~ABC\). Start with \(AD/AB~=~AE/AC\), substitute \(AB~=~AD~+~DB\) and \(AC~=~AE~+~EC\), and algebraically isolate the ratio of the \(\text{two}~\text{split}~\text{segments}\) on each side.

Unworked triangle ABC with D on AB, E on AC, and DE marked parallel to BC; segment pieces AD, DB, AE, and EC are labeled symbolically, while the split-side proportion is withheld.

Big Picture

What this problem is really about

The desired split-side ratio is hidden inside a whole-side similarity proportion. We’ll replace each whole side with the sum of its two pieces, clear the denominators, expand both products, cancel only the identical common term, and divide to expose the proportional split.

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Since △ADE ∼ △ABC, corresponding sides are proportional, so BD/DC = AB/AC and that proves the segments are split proportionally.

Since △ADE ∼ △ABC, use the geometric-mean relationship for similar triangles to conclude AD² = DB·AB, which shows AD/DB = AE/EC.

Since △ADE ∼ △ABC, AD/AB = AE/AC; replacing AB with AD + DB and AC with AE + EC gives AD/DB = AE/EC.

Since △ADE ∼ △ABC, AD/AB = AE/AC, so that proportion itself is the statement we needed to prove.

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Complete the side-splitter proof from \(\triangle~ADE~∼~\triangle~ABC\). Start with \(AD/AB~=~AE/AC\), substitute \(AB~=~AD~+~DB\) and \(AC~=~AE~+~EC\), and algebraically isolate the ratio of the \(\text{two}~\text{split}~\text{segments}\) on each side.

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