Triangles Inscribed in a Triangle

Try this

A triangle whose three vertices lie on the three sides of $△ A B C$ is called an inscribed triangle of $△ A B C$ .

First keep $D$ fixed and drag only $E$ and $F$ : the perimeter changes. Is there a clear “best spot”? It is hard to tell by feel.
Click “Unfold”. $D_{1}$ and $D_{2}$ are the reflections of $D$ in $A B$ and in $A C$ . The orange path $D_{1} \to F \to E \to D_{2}$ has exactly the same length as the perimeter of $△ D E F$ , because $F D_{1} = F D$ and $E D_{2} = E D$ .
The ends $D_{1}$ and $D_{2}$ of the path do not move when $E$ and $F$ move. The shortest path between two points is a straight segment: click “Pull E, F onto line D₁D₂” and the perimeter drops at once to the length of the green segment. Now the two angles at $F$ are equal, and so are the two at $E$ , just like light reflecting off a mirror.
Now drag $D$ . The length of the green segment $D_{1} D_{2}$ changes with it. Where should $D$ be to make it shortest? Click “Move D to the foot of the altitude” to test your guess.

Predict first

Of all the triangles inscribed in an acute triangle, which one has the shortest perimeter? What are its three vertices?

The midpoints of the three sidesThe feet of the three altitudesThe points where the incircle touches the sidesThe points where the angle bisectors meet the opposite sides

Choose an answer first; then you can see the result.

Why: reflect twice, then pull straight

The orthic triangle has the shortest perimeter

Step 1 / 6

A closed path
The perimeter of $△ D E F$ is the sum of three pieces. All three points are moving, so there is nothing obvious to compare it with.

$D E + E F + F D$
Reflect D in two mirrors
Pull it straight
How long is D₁D₂?
Make AD as short as possible
The other two points are feet too

This proof uses two ideas that will keep coming back:

Reflection turns “shortest bent path” into “the shortest path between two points is a straight segment”. It is the same trick as in Heron's classic shortest-path problem (walk to a river and then on to a camp).
The extremal principle: first fix one point and solve the problem completely, then move that point to the most extreme position (the shortest $A D$ ).

Another common inscribed triangle: the midpoint triangle

Joining the midpoints of the three sides gives the midpoint triangle. Each of its sides is a midline, parallel to the third side and half as long. So the midpoint triangle is similar to the original with ratio $\frac{1}{2}$ : its perimeter is $\frac{1}{2}$ of the original and its area is $\frac{1}{4}$ . The four small triangles are congruent.

Remove one condition

Break it

Counterexample lab: remove one condition

Conclusion: The orthic triangle is the inscribed triangle with the shortest perimeter

An acute triangle. Unfold, pull straight, move $D$ to the foot, and you get the shortest perimeter.

Practice

Exercise 1Basic

$△ A B C$ has perimeter $24$ . What is the perimeter of the triangle formed by joining the midpoints of its sides?

Exercise 2Basic

$△ A B C$ has area $36$ . What is the area of its midpoint triangle?

Exercise 3Advanced

In the acute triangle $△ A B C$ , $∠ A = 3 0^{\circ}$ . $D$ is a fixed point on $B C$ with $A D = 8$ . $E$ and $F$ move along $A C$ and $A B$ . Find the minimum perimeter of $△ D E F$ .

Exercise 4Challenge

In the acute triangle $△ A B C$ , $∠ A = 5 0^{\circ}$ , $∠ B = 6 0^{\circ}$ , $∠ C = 7 0^{\circ}$ . $D$ , $E$ , $F$ are the feet of the altitudes on $B C$ , $C A$ , $A B$ . Find $∠ E D F$ in degrees.

Try this

Why: reflect twice, then pull straight

The orthic triangle has the shortest perimeter

Another common inscribed triangle: the midpoint triangle

Remove one condition

Counterexample lab: remove one condition

Practice

Mistake Museum · Exhibits from this stop

Thinking the midpoint triangle has the shortest perimeter

Forgetting the condition “acute triangle”