Theorems on Proportional Segments in a Triangle

Try this

Drag $D$ , $E$ , $F$ anywhere: the three lines $A D$ , $B E$ , $C F$ usually enclose a small red triangle, and the product gauge is not at $1$ .
Click “Move F so CF passes through the meeting point”: the small triangle shrinks to a point and the gauge points exactly at $1$ . Now drag the vertex $A$ to change the shape of the triangle. The product does not move.
Click “3 medians”: all three ratios are $1$ , so of course the product is $1$ . That is why the three medians meet at one point.
Click “3 bisectors” and “3 altitudes”: the three ratios are all different, yet their product is still $1$ .

Ceva's theorem: if $D$ , $E$ , $F$ lie on the sides $B C$ , $C A$ , $A B$ of $△ A B C$ , then

A D, B E, C F concurrent ⟺ \frac{A F}{F B} \cdot \frac{B D}{D C} \cdot \frac{C E}{E A} = 1.

Predict first

We know $B D : D C = 2 : 1$ and $C E : E A = 3 : 1$ . For $A D$ , $B E$ , $C F$ to meet at one point, what must $\frac{A F}{F B}$ be?

6

\frac{3}{2}

\frac{1}{6}

\frac{2}{3}

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

Why: the area method again

Proof of Ceva's theorem (concurrent ⇒ product is 1)

Step 1 / 5

Let the three lines meet at P
$A D$ , $B E$ , $C F$ meet at a point $P$ . Joining $P$ to the three vertices makes three small triangles: $△ P B C$ , $△ P C A$ , $△ P A B$ .
The first ratio
The second ratio
The third ratio
Multiply

The angle bisector theorem

If $A D$ bisects $∠ B A C$ with $D$ on $B C$ , then $\frac{B D}{D C} = \frac{A B}{A C}$ : the angle bisector cuts the opposite side into two pieces in the same ratio as the two neighboring sides.

Drag

D

until the two angles are equal. What about

B D : D C

and

A B : A C

now? Then drag

A

to try other triangles.

The proof takes two lines. $D$ is on the angle bisector, so it is the same distance $h$ from $A B$ and from $A C$ . Compute the ratio of the areas of $△ A B D$ and $△ A C D$ in two ways:

\frac{S _{△ A B D}}{S _{△ A C D}} = \frac{\frac{1}{2} A B \cdot h}{\frac{1}{2} A C \cdot h} = \frac{A B}{A C} (bases A B, A C, both with height h),

\frac{S _{△ A B D}}{S _{△ A C D}} = \frac{B D}{D C} (same apex A, same height) .

The left-hand sides are the same, so $\frac{B D}{D C} = \frac{A B}{A C}$ .

Menelaus's theorem

If a straight line meets the lines of the three sides of $△ A B C$ at $F$ (on $A B$ ), $E$ (on $C A$ ) and $D$ (on the extension of $B C$ ), then again

\frac{A F}{F B} \cdot \frac{B D}{D C} \cdot \frac{C E}{E A} = 1.

Drag

E

and

F

to turn the purple line;

D

moves along with it. Does the product ever change?

Idea of the proof: through $C$ draw a line parallel to $A B$ , meeting line $F D$ at $G$ . This gives two pairs of similar triangles, with $\frac{B D}{D C} = \frac{F B}{C G}$ and $\frac{C E}{E A} = \frac{C G}{A F}$ . Substitute, and $C G$ cancels. You can also use the common-side theorem, taking two points of the line as the shared side.

Remove one condition

Break it

Counterexample lab: remove one condition

Conclusion: $\frac{A F}{F B} \cdot \frac{B D}{D C} \cdot \frac{C E}{E A} = 1$

$D$ is locked onto line $F E$ .

Practice

Exercise 1Basic

In $△ A B C$ , $A B = 6$ , $A C = 9$ , $B C = 10$ , and $A D$ bisects $∠ B A C$ , meeting $B C$ at $D$ . Find the length of $B D$ .

Exercise 2Basic

$D$ , $E$ , $F$ lie on the sides $B C$ , $C A$ , $A B$ of $△ A B C$ , and $A D$ , $B E$ , $C F$ meet at one point. Given $B D : D C = 1 : 2$ and $C E : E A = 3 : 2$ , find $\frac{A F}{F B}$ .

Exercise 3Advanced

A straight line meets sides $A B$ and $A C$ of $△ A B C$ at $F$ and $E$ , and the extension of $B C$ at $D$ . Given $A F : F B = 2 : 1$ and $C E : E A = 1 : 3$ , find $\frac{B D}{D C}$ .

Exercise 4Challenge

$A D$ is a median of $△ A B C$ , $E$ is the midpoint of $A D$ , and line $B E$ meets $A C$ at $F$ . Find $\frac{A F}{F C}$ .

Try this

Why: the area method again

Proof of Ceva's theorem (concurrent ⇒ product is 1)

The angle bisector theorem

Menelaus's theorem

Remove one condition

Counterexample lab: remove one condition

Practice

Mistake Museum · Exhibits from this stop

Concluding “collinear” because the product is 1

The three ratios do not join up end to end