Final Assignment part II
In triangle ADG, AF, DI, and GC are concurrent at point E and the following ratio is true: = 1.
This means that the product of the “left” segments divided by the product of the “right” segments equals 1 when rays extended from each vertex meet at a common point on the interior of the triangle (point E).
In order to begin, we will extend the sides of the triangle and also the rays from each vertex of the triangle. A line will be drawn through point A parallel to the opposite side GD. The extended rays from points G and D will intersect this new line at points B and H respectively.
From these new constructions similar triangles can be found
Angle ACB and angle DCG are vertical angles and therefore congruent. Angle BAC and angle CDG are alternate interior angles and also congruent. Since 2 of the three angles in these triangles are congruent then the third pair also must be congruent. Therefore ΔABC is similar to ΔDGC.
Following the same ideas ΔDIG is similar to ΔHIA (vertical angles HIA and DIG and alternate interior angles HAI and DGI)
Also ΔDEF and ΔHEA are similar for the same reasons.
Likewise, ΔFGE and ΔAEB are similar.
From these similar triangles:
ΔFGE and ΔAEB
ΔDEF and ΔHEA
ΔDIG and ΔHIA
ΔABC and ΔDGC
Many helpful ratios can be created.
= = = =
Based on these ratios, = which can be manipulated to give =
We can now substitute back into the original equation:
Click here for a Geometer’s Sketchpad file that you can manipulate.