Aniko

Use similarity inside right triangles

Lesson progressPractice problems 0/3
Difficulty
Advanced
Estimated time
28 minutes
Techniques
Right-triangle-similarityAltitude-to-hypotenuseAngle-angle-similarityGeometric-meanProportions

What you’ll learn

  1. Spot an altitude drawn from the right angle of a right triangle to its hypotenuse.
  2. Explain with AA why that altitude creates three similar right triangles.
  3. Match up vertices even when the smaller triangles are turned.
  4. Use three geometric-mean relationships to find a leg, the altitude, or a piece of the hypotenuse.
  5. Carry a result into a bigger goal, such as an area.

Prerequisites

Why this matters on the SAT

One altitude unlocks three triangles

Some SAT diagrams draw a segment from the right angle of a right triangle to the hypotenuse, meeting it at a right angle. That one segment splits the triangle into two smaller right triangles, and both are similar to the big one. Once you spot it, a short proportion can replace a long chain of calculations. Here’s a typical question.

Solution to the example

The altitude cuts the hypotenuse into two pieces, 55 and 4545. In this setup, the altitude squared equals the product of those two pieces:

CD2=AD⋅DB=5(45)=225.\begin{aligned} CD^2&=AD\cdot DB\\[1.4em] &=5(45)\\[1.4em] &=225. \end{aligned}

CDCD is a length, so take the positive square root:

CD=15.CD=15.

The answer is B. The key move came before any arithmetic: spotting an altitude from the right angle to the hypotenuse.

SAT example

The two smaller triangles are turned, but their angle marks show which vertices match. Figure not drawn to scale.

In right triangle ABCABC, ∠C=90∘\angle C=90^\circ. Altitude CD‾\overline{CD} meets hypotenuse AB‾\overline{AB} at DD. If AD=5AD=5 and DB=45DB=45, what is the length of CD‾\overline{CD}?

  1. A

    1010

  2. B

    1515

  3. C

    2525

  4. D

    5050

Recognize the three linked triangles

Here’s what to look for. All three must be true:

  1. △ABC\triangle ABC is a right triangle.
  2. A segment starts at the right-angle vertex, CC.
  3. It meets the hypotenuse, AB‾\overline{AB}, at a right angle.

A segment like CD‾\overline{CD}, drawn from a vertex and perpendicular to the opposite side, is called an altitude. Point DD sits on the hypotenuse, so the altitude splits AB‾\overline{AB} into two pieces, AD‾\overline{AD} and DB‾\overline{DB}.

When you see this setup, you get three similar triangles:

△ABC∼△ACD∼△CBD.\triangle ABC\sim\triangle ACD\sim\triangle CBD.

AA shows why.

  • △ABC\triangle ABC and △ACD\triangle ACD share ∠A\angle A, and each has a right angle: ∠C\angle C in the big triangle, ∠D\angle D in the small one. That’s two pairs of equal angles, so they’re similar.
  • △ABC\triangle ABC and △CBD\triangle CBD share ∠B\angle B, and again each has a right angle, ∠C\angle C and ∠D\angle D. So they’re similar too.

The order of the letters tells you which vertices match:

△ABC∼△ACD,A↔A,B↔C,C↔D;\begin{aligned} \triangle ABC&\sim\triangle ACD,\\[1.4em] A&\leftrightarrow A,\quad B\leftrightarrow C,\quad C\leftrightarrow D; \end{aligned}

and

△ABC∼△CBD,A↔C,B↔B,C↔D.\begin{aligned} \triangle ABC&\sim\triangle CBD,\\[1.4em] A&\leftrightarrow C,\quad B\leftrightarrow B,\quad C\leftrightarrow D. \end{aligned}

Sides match by their endpoints. In the first pair, A↔AA\leftrightarrow A and C↔DC\leftrightarrow D, so side AC‾\overline{AC} of the big triangle matches side AD‾\overline{AD} of the small one. In a diagram, the smaller triangles may be turned or flipped, and that’s what makes this matching the tricky part.

Common mistake:

Matching sides by where they sit on the page. The small triangles are turned, so “the left side” of one isn’t the left side of another. Mark the shared acute angle and the two right angles first, write the vertices in matching order, and only then pair up sides by their endpoints.

Check your understanding:

Which two pairs of equal angles show that △ABC∼△ACD\triangle ABC\sim\triangle ACD by AA?

Build the geometric-mean proportions

The geometric mean of two positive lengths is the length whose square equals their product. For 44 and 99, it’s 66, because 62=36=4⋅96^2=36=4\cdot9. In general, if xx is the geometric mean of uu and vv, then

x2=uvandx=uv.x^2=uv \qquad\text{and}\qquad x=\sqrt{uv}.

The three similar triangles give you three relationships like this, one for the altitude and one for each leg.

The altitude

The altitude is the geometric mean of the two pieces of the hypotenuse. That’s the one you used in the SAT example:

CD2=AD⋅DB.\boxed{CD^2=AD\cdot DB}.

The leg next to ADAD

Leg AC‾\overline{AC} touches the piece AD‾\overline{AD}. It’s the geometric mean of the whole hypotenuse and that piece:

AC2=AB⋅AD.\boxed{AC^2=AB\cdot AD}.

The leg next to DBDB

Leg BC‾\overline{BC} works the same way with the piece it touches, DB‾\overline{DB}:

BC2=AB⋅DB.\boxed{BC^2=AB\cdot DB}.

You don’t have to memorize these as three separate facts. Each one comes from matching sides of two similar triangles. In △ABC∼△ACD\triangle ABC\sim\triangle ACD, side ACAC matches ADAD, and side ABAB matches ACAC, so

ACAB=ADAC.\frac{AC}{AB}=\frac{AD}{AC}.

Cross-multiply and you get AC2=AB⋅ADAC^2=AB\cdot AD. ACAC belongs to both triangles, which is why it ends up squared. Matching sides of the two smaller triangles the same way gives

CDAD=DBCD,\frac{CD}{AD}=\frac{DB}{CD},

so CD2=AD⋅DBCD^2=AD\cdot DB.

Here’s the short version to remember: the altitude multiplies the two pieces, and a leg multiplies the whole by its own piece. The whole hypotenuse is just the two pieces added together:

AB=AD+DB.AB=AD+DB.
Try it yourself:

Without calculating, pick the relationship you’d use to find leg ACAC when you know ABAB and ADAD. Then explain why DBDB doesn’t belong in it.

Answer: Use AC2=AB⋅ADAC^2=AB\cdot AD. ADAD is the piece next to leg ACAC. DBDB sits next to the other leg, BCBC.

Common mistake:

Writing AC2=AD⋅DBAC^2=AD\cdot DB gives a leg the altitude’s product. Only the altitude multiplies the two pieces. So name what you’re finding first: for leg ACAC, it’s the whole hypotenuse times the piece ACAC touches, AC2=AB⋅ADAC^2=AB\cdot AD.

Example: Find a leg through two linked proportions

Worked example

Use the altitude to find the missing piece of the hypotenuse first. Then use the leg relationship that pairs BCBC with the piece it touches.

In right triangle ABCABC, ∠C=90∘\angle C=90^\circ, and altitude CD‾\overline{CD} meets hypotenuse AB‾\overline{AB} at DD. If CD=10CD=10 and AD=4AD=4, which expression gives the length of BC‾\overline{BC} in simplest radical form?

  1. A

    725\sqrt{725}

  2. B

    5295\sqrt{29}

  3. C

    2525

  4. D

    2292\sqrt{29}

Step 1

Check the setup

The altitude starts at the right angle and meets the hypotenuse, so this is the setup with three similar triangles:

△ABC∼△ACD∼△CBD.\triangle ABC\sim\triangle ACD\sim\triangle CBD.

That means the geometric-mean relationships work here.

Step 2

Find the missing piece

You want BCBC, and its relationship needs ABAB and DBDB. You don’t have either one yet. The altitude gets you there, because it ties the two pieces together:

CD2=AD⋅DB102=4(DB)100=4DBDB=25.\begin{aligned} CD^2&=AD\cdot DB\\[1.4em] 10^2&=4(DB)\\[1.4em] 100&=4DB\\[1.4em] DB&=25. \end{aligned}

Now add the pieces to get the whole hypotenuse:

AB=AD+DB=4+25=29.AB=AD+DB=4+25=29.

Step 3

Match the leg to its piece

Leg BC‾\overline{BC} touches the piece DB‾\overline{DB}, so

BC2=AB⋅DB.BC^2=AB\cdot DB.

Use the whole hypotenuse, 2929, and the piece next to BCBC, 2525:

BC2=29(25).BC^2=29(25).

That product isn’t a quick one by hand, so enter 29*25 in Desmos. It gives 725725.

Step 4

Keep the answer exact

Take the positive square root, then pull out the perfect square 2525:

BC=725=25⋅29=529.\begin{aligned} BC&=\sqrt{725}\\[1.4em] &=\sqrt{25\cdot29}\\[1.4em] &=5\sqrt{29}. \end{aligned}

So

BC=529.\boxed{BC=5\sqrt{29}}.

The answer is B. The question asks for simplest radical form, so you simplify the root rather than round it. Choice A is the same length, just not simplified.

Common mistake:

It’s tempting to write 102+BC2=AB210^2+BC^2=AB^2, as if CD=10CD=10 were a leg of the big triangle. But CDCD runs inside the big triangle, so it isn’t one of its sides. Use CD2=AD⋅DBCD^2=AD\cdot DB first, then the relationship for the leg you want.

Check your understanding:

Using the same diagram and numbers, what is the exact length of ACAC?

Choose the relationship from your target

Before you write an equation, give each length its job: the whole hypotenuse, one of its two pieces, the altitude, or a leg, along with the piece that leg touches. The letters change from problem to problem, so go by job, not by letter.

Then pick the relationship that holds what you want and what you know. If something’s missing, you may need three steps, as in the worked example:

  1. Use CD2=AD⋅DBCD^2=AD\cdot DB to find a missing piece.
  2. Add AD+DBAD+DB to get the whole hypotenuse.
  3. Use the relationship for the leg you want.

Two triangles that share angles aren’t the signal on their own. That’s an ordinary similar-triangle question. The signal here is more specific: an altitude from the right angle to the hypotenuse. And when you already know two sides of one right triangle, that’s a direct Pythagorean problem instead.

Desmos can’t pick the relationship for you. That comes from reading the diagram. Once you’ve picked it, do quick products by hand, and simplify exact radicals by hand too. When the product or root isn’t quick, let Desmos do that arithmetic in one line. For x2=uvx^2=uv, type sqrt(u*v) with the real numbers in place of uu and vv, instead of working it out in several stages.

Practice problems

These start with the altitude, move to a leg, and finish with an area. For each one, say what you’re finding before you choose a relationship.

Find the altitude

Practice problem

In right triangle JKLJKL, ∠K=90∘\angle K=90^\circ. Altitude KM‾\overline{KM} meets hypotenuse JL‾\overline{JL} at MM. If JM=6JM=6 and ML=35ML=35, what is the length of KM‾\overline{KM}, to the nearest tenth?

Answer choices
Calculator loads as you approach
Pick the altitude’s relationship first. Then enter sqrt(6*35) for the root.

Match a leg to its piece

Practice problem

In right triangle RSTRST, ∠S=90∘\angle S=90^\circ. Altitude SU‾\overline{SU} meets hypotenuse RT‾\overline{RT} at UU. If RU=9RU=9 and UT=7UT=7, what is the length of RS‾\overline{RS}?

Calculator loads as you approach
Find the whole hypotenuse and the piece next to RSRS before you calculate.

Carry similarity into area

Practice problem

In right triangle ABCABC, ∠C=90∘\angle C=90^\circ. Altitude CD‾\overline{CD} meets hypotenuse AB‾\overline{AB} at DD. The ratio AD:DBAD:DB is 4:94:9, and CD=30CD=30. What is the area, in square units, of triangle ABCABC?

Calculator loads as you approach
Use the ratio and the altitude to find the hypotenuse, then find the area.

Finish the lesson

3 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

  • Look for an altitude from the right-angle vertex to the hypotenuse.
  • It creates △ABC∼△ACD∼△CBD\triangle ABC\sim\triangle ACD\sim\triangle CBD. Each small triangle shares an acute angle with the big one, and each has a right angle, so AA applies.
  • Match vertices by equal angles, not by how the triangles are turned.
  • The altitude multiplies the two pieces: CD2=AD⋅DBCD^2=AD\cdot DB.
  • A leg multiplies the whole hypotenuse by the piece it touches, like AC2=AB⋅ADAC^2=AB\cdot AD.
  • Add AD+DBAD+DB to get ABAB, and keep going until you reach what the question asks for.
  • Pick the relationship from the diagram. Use Desmos for a product or root that isn’t quick, but keep exact radicals when the question asks for them.

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