The Simson Line and its applications

If perpendiculars are dropped onto the sides of a triangle ABC from a point P, the feet of these perpendiculars usually form the vertices of a new triangle A₁ B₁ C₁ (so called the pedal triangle of P). Let us now examine the exceptional case where the point P lies on the circumcircle, as in Figure 4
Problem 4. The feet of the perpendiculars from a point to the sides of a triangle are collinear if and only if the point lies on the circumcircle. (The line containing the feet is known as the Simson line).

  

Figure 4: The Simson line

Proof. Let first the point is on the circle. Using the already introduced notations, without loss of generality we can take the point P to lie on the arc CA that does not contain B and at least as far from C as from A (see Figure 4). All the other cases can be derived by renaming A, B and C. Because of the right angles at A₁, B₁ and C₁, P lies also on the circumcircles of the triangles $\Delta A_1 B C_1$, A₁ B₁ C and AB₁ C₁. Therefore

$$\angle APC = 180^{\circ} - \angle B = \angle C_1 P A_1$$

and, subtracting $\angle APA_1$, we deduce $\angle A_1 PC = \angle C_1 PA$. But since the points A₁, C, P and B₁ lie on a circle, $\angle A_1 PC = \angle A_1 B_1 C$, and since the points A, B₁, P and C₁ lie on a circle, $\angle C_1 PA = \angle C_1 B_1 A$. Thus

$$\angle A_1 B_1 C = \angle C_1 B_1 A,$$

so the points are collinear.

Conversely, if a point P is so situated that the pedal triangle A₁ B₁ C₁ is degenerate, P must evidently lie in the region of the plane that is inside one angle of $\Delta ABC$ and beyond the opposite side, By re-naming the vertices if necessary, we can assume that this ``one angle'' is B, and that C₁ lies on the extension of the side BA beyond A, as in Figure 4. We can then reverse the steps in the above discussion of angles and conclude that $\angle APC = 180^{\circ} - \angle B$, so P lies on the circumcircle.

Problem 4. In the case of Figure 4 prove that

$$\frac{AC}{PB_1}=\frac{BC}{PA_1}+\frac{AB}{PC_1}.$$

Proof. Introducing the following notations: $\angle ABP = \alpha$, $\angle BCP = \beta$ and $\angle CAP = \gamma$ we have that $\angle ACP = \alpha$, $\angle BAP = 180^{\circ} - \beta$ and $\angle CBP = \gamma$. So $PA_1 = PB \sin \gamma = 2R\sin \beta \sin \gamma$ (we used the Law of Sines and that the circumcircles of $\Delta ABC$ and $\Delta PBC$ are the same so they have the same radius). Similarly $PB_1 = 2R\sin \alpha \sin \gamma$ and $PC_1 = 2R\sin \alpha \sin \beta$. Also we have that $AB=2R \sin \angle APB = 2R \sin (180^{\circ} - \angle BAP - \angle ABP) = 2R\sin (\beta - \alpha)$. Similarly we have that $BC=2R\sin (\beta + \gamma)$ and $AC= 2R\sin (\alpha + \gamma)$. So we have to verify the following identity:

$$\frac{ \sin (\alpha + \gamma)}{\sin \alpha \sin \gamma}=\frac... ... \gamma}+\frac{ \sin (\beta - \alpha)}{\sin \alpha \sin \beta},$$

which is easy.

The concept of the Simson line can be used to derive a very useful theorem, as follows.

Ptolemy's Theorem. If a quadrilateral is inscribed in a circle, the sum of the products of the two pairs of opposite sides is equal to the product of the diagonals.

Proof. Using again Figure 4 and its notations we can take the quadrilateral to be ABCP. The sides of the pedal triangle (in our case degenerate) A₁ B₁ C₁ can be found easily. Applying the Law of Sines to $\Delta AB_1 C_1$ and $\Delta ABC$ we obtain

$$\frac{B_1C_1}{\sin \angle A}=AP, \frac{BC}{\sin \angle A}=2R$$

because the right angles at B₁ and C₁ indicate that these points lie on the circle with diameter AP. SO using the standard notation for the sides of the triangle ABC we find that

$$B_1C_1=a\frac{AP}{2R},C_1A_1=b\frac{BP}{2R},A_1B_1=c\frac{CP}{2R}.$$

Since A₁B₁+B₁C₁=A₁C₁, we deduce that cCP+aAP=bBP, that is

AB.CP+BC.AP=AC.BP,

so we proved the Ptolemy's Theorem.