## 1. It follows from the construction of the parallel line that angle AFC is congruent to angle HPC, and angle FAC is congruent to angle PHC (by corresponding angles). This implies that triangle AFC is similar to triangle HPC. Since triangle AFC is similar to triangle HPC, this implies that .

## 2. We also know that angle CFB is congurent to angle CPI and angle CBF is congruent to angle CIP (by corresponding angles). This implies that triangle CFB is similar to triangle CPI. It follows that .

## 3. From 1 and 2 we have , transitive property of equality. We could also write this as: .

## 4. Also from the construction, we have angle ABE=angle HEP. We also know, by corresponding angles that angle ABE is congruent to HPE. This implies that triangle ABE is similar to triangle HPE, by angle-angle similarity. It follows that .

## 5. Also from the construction, we have angle ADB=angle PDI. We also know, by corresponding angles that angle BAD is congruent to IPD. This implies that triangle BAD is similar to triangle IPD, by angle-angle similarity. It follows that .

## .

## 1. Through similar arguments as before, we will see that triangle ABD is similar to triangle AJP and triangle DAC is similar to triangle PAK. These yield the following equalities:

## This implies , which implies .

## 2. By the same argument, triangle BCE is similar to triangle PKE. This gives us the following equality:

. ## 3. By the same previous argument, triangle BCF is similar to triangle JPF. This implies that .

## 1. With this construction, we could show as before that the following triangles are similar and thus yield the following ratios:

## a. triangle BPM is similar to triangle BEC

## b. triangle ABE is similar to triangle LBP

## a&b imply that i.e. .

## c. triangle ACD is similar to triangle PMD

## d. triangle ACF is similar to triangle LFB

Click here to see a working GSP sketch.

## Through my investigation, I found that the ratio of the areas of triangle ABC and triangle DEF is equal to four when F, E, and D are the midpoints of their corresponding sides. We know this is true because triangle DEF forms a medial triangle. The area of the medial triangle is always 1/4 of the area of the original triangle, in our case triangle DEF is 1/4 the area of triangle ABC.