Parts of Similar Triangles

Q: Does this work if the triangles are congruent?

Congruent triangles are similar with scale factor $1$, so yes — corresponding altitudes, medians, and bisectors are simply equal. The proportion still holds; it just says the two segments have the same length.

When two triangles are similar, the scale factor does not stop at the sides. Every corresponding length inside the triangles scales by it too — altitudes, medians, angle bisectors, even perimeters. One ratio rules them all.

That single fact turns a whole family of problems into the same proportion. Find the ratio of corresponding sides, set it equal to the ratio of the special segments, cross multiply, done.

The three special segments

A median runs from a vertex to the midpoint of the opposite side. An altitude runs from a vertex and meets the opposite side at a right angle. An angle bisector cuts the vertex angle in half on its way to the opposite side.

In diagrams, altitudes are usually drawn dashed with a right-angle mark at the foot, and medians land on a side marked with matching ticks. Identify which segment you are looking at first — the setup is the same either way, but you need the corresponding pair.

One ratio rules everything

If two triangles are similar with scale factor $k$ , then corresponding altitudes are in the ratio $k$ , corresponding medians are in the ratio $k$ , and corresponding angle bisectors are in the ratio $k$ . In symbols: $\dfrac{\text{altitude of } \triangle ABC}{\text{altitude of } \triangle DEF} = \dfrac{AB}{DE}$ .

The trap to avoid: the ratio is not squared. Squaring the scale factor is for areas. Altitudes, medians, and bisectors are lengths, so they scale exactly like the sides. Sides in the ratio $3:5$ means medians in the ratio $3:5$ — never $9:25$ .

Setting up the proportion

Step 1: find the scale factor from a pair of corresponding sides. Step 2: set the ratio of the special segments equal to it. Step 3: cross multiply and solve.

Make sure the segments actually correspond — the altitude from the smallest angle of one triangle pairs with the altitude from the smallest angle of the other. Keep the same triangle on top in both fractions and the algebra takes care of itself.

Worked examples

Example 1: corresponding altitudes

Two similar triangles have corresponding sides $18$ and $12$ . The altitude of the larger triangle is $8$ . Find the corresponding altitude $x$ of the smaller triangle.

Altitudes scale like the sides

\dfrac{x}{8} = \dfrac{12}{18}

Cross multiply

18x = 96

Divide both sides by

18

x = \dfrac{96}{18} = \dfrac{16}{3}

Answer: $x = \dfrac{16}{3}$

Example 2: corresponding medians

$\triangle JKL \sim \triangle PQR$ with $KL = 18$ and $QR = 12$ . A median of $\triangle JKL$ measures $15$ . Find the corresponding median of $\triangle PQR$ .

Medians scale like the sides

\dfrac{15}{m} = \dfrac{18}{12}

Cross multiply

18m = 180

Divide both sides by

18

m = 10

Answer: The corresponding median measures $10$

Example 3: an angle bisector with a scale factor

$\triangle ABC \sim \triangle DEF$ with scale factor $\dfrac{4}{3}$ from $\triangle ABC$ to $\triangle DEF$ ... wait, from $\triangle DEF$ to $\triangle ABC$ . An angle bisector of $\triangle ABC$ is $20$ units long. Find the corresponding angle bisector of $\triangle DEF$ .

Bisectors scale like the sides

\dfrac{20}{b} = \dfrac{4}{3}

Cross multiply

4b = 60

Divide both sides by

4

b = 15

Answer: $b = 15$

Try one yourself

Common questions

Is the ratio of the medians the scale factor squared?

No. Squaring is only for areas. A median is a length, so corresponding medians are in the same ratio as corresponding sides. Sides in the ratio $3:5$ means medians in the ratio $3:5$ , while the areas would be in the ratio $9:25$ .

Do altitudes, medians, and angle bisectors all use the same ratio?

Yes. Any pair of corresponding lengths in similar triangles — the three special segments, the perimeters, even corresponding halves of sides — scales by the one scale factor. Only area behaves differently.

How do I know which altitude corresponds to which?

Corresponding altitudes drop from corresponding vertices. Use the similarity statement: in $\triangle ABC \sim \triangle DEF$ , the altitude from $A$ pairs with the altitude from $D$ . In a figure, match them by the sides they land on.

Does this work if the triangles are congruent?

Congruent triangles are similar with scale factor $1$ , so yes — corresponding altitudes, medians, and bisectors are simply equal. The proportion still holds; it just says the two segments have the same length.

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Parts of Similar Triangles

The three special segments

One ratio rules everything

Setting up the proportion

Worked examples

Example 1: corresponding altitudes

Example 2: corresponding medians

Example 3: an angle bisector with a scale factor

Try one yourself

Common questions

Keep learning

Similar Polygons & Scale Factor

SSS & SAS Similarity

The Triangle Midsegment Theorem

Similarity Proofs & Indirect Measurement