# A geometric inequality

#### …and its solution by Dan Hoey

Drag the point $P$. Press “r” to reset the diagram to its initial state. Proposition: $PC \le PA + PB$.

## The problem

Proposition: Let $ABC$ be an equilateral triangle and $P$ be an arbitrary point in its plane. Then $PC \le PA + PB$.

This was brought up in a message on the `geometry.puzzles` newsgroup on November 11, 2001. Go to that message and scroll to the bottom of the page to see the discussion thread.

On November 22, 2001 Dan Hoey offered a particularly nice solution. He also commented that he had learned that problem may be related to the Van Schooten Theorem, which indeed it is. See Van Schooten's and Pompeiu's Theorems: What are these? for much detail and historical background.

## The proof

Here is Dan Hoey's proof of the proposition as stated above.

On the line segment $AP$ construct the equilateral triangle $APD$, as shown in the diagram below, then add the line segment $DC$.

Let us show that the triangles $APB$ and $ADC$ are congruent. For this, Let us observe that the sides $AP$ and $AB$ in the triangle $APB$ equal the sides $AD$ and $AC$ in the triangle $ADC$, by the construction. Moreover, the angles $BAP$ and $CAD$ are equal because each equals the difference of a 60 degree angle and the angle $DAB$. Therefore, the triangles $APB$ and $ADC$ are congruent by the side-angle-side equality. We conclude, in particular, that $PB = DC$.

In the triangle $PDC$ we have $PC \le PD + DC$. In this inequality replace $PD$ and $DC$ by their equivalents $PA$ and $PB$ to arrive at $PC \le PA + PB$. QED

Drag the point $P$. Press “r” to reset the diagram to its initial state. Observation: The triangles $APB$ and $ADC$ are congruent.

This applet was created by Rouben Rostamian using David Joyce's Geometry Applet on November 23, 2001.
Cosmetic revisions on June 23, 2010.

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