Ratio of the two splitted fragments created by drawing a perpendicular projection from one vertex (i.e., one angle) of a triangle to its opossite side

[sankar basu]

In triangle ABC, Let D be the foot of the perpendicular drawn from B onto the opposite side AC. Then, is it that the ratio's AB / BC = AD / DC = <ABD/<DBC ?

I presume it is true. But, can one suggest an analytical proof of this theorem ? Or state otherwise the reason (if it is not true).

It would be of much help.

 

- Thanks and regards

Sankar Basu

Post Doc

Bioinformatics

Linkoping University, Sweden

 

 

[Logophobic]

Your presumption is wrong. Here is a counterexample:

AB/BC = 0.75
AD/DC = 0.5625