|This is the write-up of the Final Assignment||
Brian R. Lawler
Back to Final Write-up cover page.
Conjecture? Prove it! (you may need draw some parallel lines to produce some similar triangles) Also, it probably helps to consider the ratio
Can the result be generalized (using lines rather than segments to construct ABC) so that point P can be outside the triangle? Show a working GSP sketch.
The ratio of the products of the partial side lengths is equal to one. In other words:
By adding lines parallel to CP through A and B, several similar triangles are created. Namely,
With this established, a close inspection of the ratios of corresponding sides yields the desired result.
It follows that , or .
Next, notice that and .
Multiplying these fractions shows that .
Each term in the numerator of the right side cancels with the same term in the denominator, therefore the fraction on the right is equal to 1 and it must be that .
The above result holds true even when point P lies on or outside triangle ABC as well. In other words, this is a relationship that holds between any four points on a plane.
The image at right is one such example of P outside the triangle formed by A, B, and C. Click the image to explore a Geometer's Sketchpad file that will demonstrate the relationship holds.
|Comments? Questions? e-mail me at firstname.lastname@example.org|
|Last revised: January 3, 2001||