Similarity in Right Triangles

Watch the video explainer

Learning goals

Locate the altitude foot on the interior of a right triangle's hypotenuse
Justify the ordered similarity of all three triangles using AA
Derive the geometric-mean relationships for the altitude and both legs
Find missing lengths while distinguishing a whole hypotenuse from its pieces

A perpendicular cut

In a right triangle, the hypotenuse is the side opposite the right angle. The other two sides are the legs. An altitude is a perpendicular segment from a vertex to the line containing the opposite side. Its meeting point with that line is its foot.

We will use nondegenerate $\triangle ABC$ , right at $C$ . Draw altitude $CH$ to hypotenuse $AB$ . Use these length names throughout:

\begin{aligned} a&=BC\\ b&=CA\\ c&=AB\\ p&=AH\\ q&=HB\\ h&=CH. \end{aligned}

The right-angle marks identify the given triangle and its altitude.

Here $H$ is strictly between $A$ and $B$ . To see why, recall that both endpoint angles $A$ and $B$ are acute. If $H$ were outside the segment, the small triangle at the nearer endpoint would contain both a right angle at $H$ and an obtuse angle supplementary to that acute endpoint angle. Its angle sum would exceed $180^\circ$ . If $H$ were an endpoint, that endpoint would itself be a second right angle in $ABC$ . Both possibilities are impossible.

Thus $p,q,h$ are positive, and the two pieces make the whole:

c=p+q.

The foot need not be the midpoint. Perpendicularity supplies right angles; it does not supply equal lengths.

Check your understanding

Why does the altitude from the right-angle vertex meet the open hypotenuse segment?

A An altitude always bisects its opposite side B Both hypotenuse endpoint angles are acute, ruling out an exterior or endpoint foot C All right triangles have equal legs D The altitude must be as long as a leg

Keep the three triangles in order

The cut preserves enough angles to make all three triangles similar. The AA criterion needs two matching angle pairs; one pair of right angles alone is not enough.

Three triangles from one altitude#

First suppose angle $A$ is $32^\circ$ and angle $B$ is $58^\circ$ . Triangle $ACH$ has the same $32^\circ$ angle at $A$ and a $90^\circ$ angle at $H$ , so its angle at $C$ is $58^\circ$ . Triangle $CBH$ has the same $58^\circ$ angle at $B$ and a $90^\circ$ angle at $H$ , so its angle at $C$ is $32^\circ$ .

Writing each triangle in the angle order $32^\circ$ , $58^\circ$ , $90^\circ$ gives

\triangle ABC\sim\triangle ACH\sim\triangle CBH.

The particular acute angles did not matter. In general, call the angle at $A$ $\alpha$ and the angle at $B$ $\beta$ . Their sum is $90^\circ$ . The same shared-angle and right-angle argument gives this correspondence:

Position	Whole triangle $ABC$	Left triangle $ACH$	Right triangle $CBH$
Angle $\alpha$	$A$	$A$	$C$
Angle $\beta$	$B$	$C$	$B$
Right angle	$C$	$H$	$H$

Each small triangle shares an acute angle with the whole triangle and has a matching right angle. AA proves each similarity. The table also gives two angle matches between the small triangles, so AA proves their similarity directly.

The two angles at $C$ in the small triangles have different names and usually different measures. The common point $C$ does not have to match itself.

Match angles after a perpendicular cut

Scroll sideways to compare the triangles.

Leg a (BC) Leg b (CA) Construction Common turn

Leg a (BC) is 6 units long. Leg b (CA) is 5 units long. Hypotenuse c (AB) is about 7.81 units long. Piece p (AH) is about 3.2 units long. Piece q (HB) is about 4.61 units long. Altitude h (CH) is about 3.84 units long. CH is perpendicular to AB at the interior point H. The angle order is ABC, ACH, and CBH. One-arc angles measure about 50.2°. Two-arc angles measure about 39.8°. The common extra turn is 0° counterclockwise.

Use the construction stages to compare the angle marks; the whole triangle and any displayed copies use the same length scale.

Worked example 1 Angles at the cut

Problem. Triangle $ABC$ is right at $C$ , $H$ lies on $AB$ , and $CH\perp AB$ . The angle at $A$ measures $41^\circ$ .

(a) Find $\angle ACH$ and $\angle HCB$ .

(b) Write all three triangles in the same corresponding order.

Only the given right angles are marked.

Solution. (a) Triangle $ACH$ contains the given angle at $A$ and the right angle at $H$ .

\begin{aligned} m\angle ACH&=180^\circ-90^\circ-41^\circ\\ m\angle ACH&=49^\circ. \end{aligned}

The whole angle at $C$ is $90^\circ$ , so

\begin{aligned} m\angle HCB&=90^\circ-49^\circ\\ m\angle HCB&=41^\circ. \end{aligned}

(b) Read each triangle in the order $41^\circ$ , $49^\circ$ , $90^\circ$ :

\triangle ABC\sim\triangle ACH\sim\triangle CBH.

Check your understanding

In $\triangle ABC\sim\triangle ACH$ , which side of the small triangle corresponds to the whole hypotenuse $AB$ ?

AH

CH

BH

AC

The altitude and the two pieces

The altitude lies between the two pieces in a repeated ratio. This gives a multiplication rule, rather than a rule about adding lengths.

The altitude geometric-mean relationship#

Take hypotenuse pieces $4$ and $9$ . From the ordered small-triangle similarity $ACH\sim CBH$ , side $AH$ matches $CH$ , while $CH$ matches $BH$ . Therefore

\frac4h=\frac h9.

Multiply by $9h$ , which is positive.

\begin{aligned} h^2&=36\\ h&=6. \end{aligned}

The labeled lengths satisfy the corresponding small-triangle ratios.

Nothing depended on the chosen pieces. With positive lengths $p$ and $q$ , the same correspondence gives

\begin{aligned} \frac ph&=\frac hq\\ h^2&=pq\\ h&=\sqrt{pq}. \end{aligned}

The geometric mean of two positive numbers is the positive square root of their product. For $4$ and $9$ , it is $6$ , since $4/6$ and $6/9$ are equal. It is different from their arithmetic mean, $(4+9)/2$ .

The altitude to the hypotenuse is the geometric mean of the two hypotenuse pieces. This requires a right triangle and the perpendicular from its right-angle vertex to its hypotenuse.

Worked example 2 Find the inner length

Problem. In the standard right-triangle configuration,

\begin{aligned} AH&=3\text{ cm}\\ HB&=12\text{ cm}. \end{aligned}

Find $CH$ exactly.

Solution. The two given lengths are the hypotenuse pieces, so the altitude relation uses their product.

\begin{aligned} CH^2&=3\cdot12\\ CH^2&=36\\ CH&=6\text{ cm}. \end{aligned}

The completed configuration for this example.

The positive root is required because $CH$ is a length.

Check your understanding

The hypotenuse pieces are $2$ cm and $7$ cm. What is the exact altitude?

\sqrt{14}

cm B

4.5

cm C

14

cm D

3

Each leg uses the whole hypotenuse

A leg connects to one endpoint of the hypotenuse. Its geometric-mean relation uses the whole hypotenuse and the piece touching that endpoint.

The two leg geometric-mean relationships#

For a numerical instance, take a whole hypotenuse of $20$ , with pieces $5$ at $A$ and $15$ at $B$ . The similarity $ABC\sim ACH$ pairs $AB$ with $AC$ and $AC$ with $AH$ .

\begin{aligned} \frac{20}{b}&=\frac b5\\ b^2&=100\\ b&=10. \end{aligned}

The similarity $ABC\sim CBH$ pairs $AB$ with $CB$ and $BC$ with $BH$ .

\begin{aligned} \frac{20}{a}&=\frac a{15}\\ a^2&=300\\ a&=10\sqrt3. \end{aligned}

Each leg is paired with the hypotenuse piece at its own endpoint.

For arbitrary positive lengths, the same two correspondences give

\begin{aligned} \frac cb&=\frac bp\\ b^2&=cp \end{aligned}

and

\begin{aligned} \frac ca&=\frac aq\\ a^2&=cq. \end{aligned}

Thus $CA$ uses $AH$ , while $BC$ uses $HB$ . The whole length $c$ occurs in both leg relations.

Worked example 3 Recover both outer lengths

Problem. In the standard configuration,

\begin{aligned} AB&=18\text{ cm}\\ AH&=8\text{ cm}. \end{aligned}

(a) Find $HB$ .

(b) Find both legs $CA$ and $BC$ exactly.

Solution. (a) Subtract the known piece from the whole.

\begin{aligned} HB&=18-8\\ HB&=10\text{ cm}. \end{aligned}

(b) The leg at $A$ uses piece $AH$ .

\begin{aligned} CA^2&=18\cdot8\\ CA&=12\text{ cm}. \end{aligned}

The leg at $B$ uses piece $HB$ .

\begin{aligned} BC^2&=18\cdot10\\ BC&=6\sqrt5\text{ cm}. \end{aligned}

The two legs are labeled after their separate calculations.

Work backward from an altitude

Changing the triangle’s orientation does not change the correspondences. Identify the right angle, hypotenuse, and foot before choosing a relation.

Worked example 4 Recover the other piece

Problem. Triangle $EFG$ is right at $F$ . The perpendicular $FT$ meets the interior of hypotenuse $EG$ at $T$ . The lengths are

\begin{aligned} ET&=6\text{ cm}\\ FT&=9\text{ cm}. \end{aligned}

(a) Find $TG$ .

(b) Find the whole hypotenuse $EG$ .

Solution. (a) Here the altitude is $FT$ , and its pieces are $ET$ and $TG$ .

\begin{aligned} FT^2&=ET\cdot TG\\ 81&=6TG\\ TG&=13.5\text{ cm}. \end{aligned}

(b) Add the pieces, rather than using either one as the whole.

\begin{aligned} EG&=6+13.5\\ EG&=19.5\text{ cm}. \end{aligned}

The completed triangle has its right-angle vertex below the hypotenuse.

Check your understanding

Triangle $ABC$ is right at $C$ , with altitude $CH$ to $AB$ . Find $AH$ from the givens.

$AB=25$

$CA=10$

10

\sqrt{250}

4

21

Similarity in Right Triangles

Learning goals

A perpendicular cut

Keep the three triangles in order

The altitude and the two pieces

Each leg uses the whole hypotenuse

Work backward from an altitude

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

Your progress