Proof of the Ratio of Areas of Similar Triangles

How are the areas of two similar triangles related to the square of the corresponding angle bisector segment? Let's prove it step by step.


1. Setting Up the Triangles

Consider two similar triangles, ΔABC and ΔDEF, where ∠A = ∠D, ∠B = ∠E, and ∠C = ∠F. Let AD and BE be the angle bisectors of these triangles, intersecting at point O.


2. Finding the Ratios of the Areas

Since ΔABC ~ ΔDEF, we have:

AB/DE = AC/DF = BC/EF

Now, let's focus on the ratio of the areas of these triangles. The area of a triangle is given by the formula:

Area = 1/2 * base * height


3. Applying the Area Formula

For ΔABC, the height corresponding to base BC is the length of the angle bisector AD, which we can denote as h1. So, the area of ΔABC is:

Area(ΔABC) = 1/2 * BC * h1

Similarly, for ΔDEF, the height corresponding to base EF is the length of the angle bisector BE, which we can denote as h2. So, the area of ΔDEF is:

Area(ΔDEF) = 1/2 * EF * h2


4. Comparing the Ratios of the Areas

Now, we need to compare the ratio of the areas of ΔABC and ΔDEF:

Area(ΔABC)/Area(ΔDEF) = (1/2 * BC * h1) / (1/2 * EF * h2)

Simplifying this expression, we get:

Area(ΔABC)/Area(ΔDEF) = (BC/EF) * (h1/h2)

Since BC/EF = AB/DE (from the similarity of the triangles), we have:

Area(ΔABC)/Area(ΔDEF) = (AB/DE) * (h1/h2)


5. Conclusion

From the above derivation, we can see that the ratio of the areas of two similar triangles is indeed equal to the square of the corresponding angle bisector segments. This relationship holds true for any pair of similar triangles with angle bisectors intersecting at a point.