This argument is long and messy, but it works. I haven't been able to find a shorter proof.

Suppose for convenience that the sides of the triangle are 1 unit in length. The first thing to notice is that the sum of the distances from P to the three sides is equal to √3/2.

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In this picture (gacked from here, where there is a full explanation), the area of the whole triangle is (1/2)base×height, and the height is √3/2. But this is equal to the sum of the three coloured triangles, whose bases are all equal to 1 and whose heights are and . So .

Notice that the area of the triangle ABC is (1/2)base×height = √3/4.

What we want to show here is that in the next picture (cunningly Photoshopped by me) the sum of the areas of the darker-coloured triangles is the same as the sum of the areas of the lighter-coloured triangles.

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Call the bases of the darker-coloured triangles and . Then their total area is .

The first main step is to find expressions for and in terms of and .

For , the vertical height of the point E above the horizontal line BC is . But it is also equal to . That leads to the equation .

Similarly, the vertical height of F above the line BC is equal to √3/2 (the height of A) minus the vertical drop from A to F. It is also equal to the drop from F to P plus the drop from P to D. This leads to the equation .

For , the distance DC is equal to and also to . This gives the equation .

Let A denote the sum of the areas of the darker-coloured triangles. Then

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Now use the fact that to write this as

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which is the area of the triangle ABC. Thus the sum of the areas of the darker-coloured triangles is half the area of the whole triangle, which is what we wanted to show.