an angle bisector of a triangle divides the opposite side of the triangle into segments 6 cm and 5cm long. A second side of the
triangle is 6.9 cm long. Find the longest and shortest possible lengths of the third side of the triangle. Round answers to the nearest tenth of a centimeter.
There is a not so well-known theorem that solves this problem.
The theorem is stated as follows: "Each angle bisector of a triangle divides the opposite side into segments proportional in length to the adjacent sides" (Coxeter & Greitzer)
This means that for a triangle ABC, where angle A has a bisector AD such that D is on the side BC, then BD/DC=AB/AC
Here either BD/DC=6/5=AB/AC, where AB=6.9, then we solve for AC=AB*5/6=5.75,
or
BD/DC=6/5=AB/AC, where AC=6.9, then we solve for AB=AC*6/5=8.28
Hence, the longest and shortest possible lengths of the third side are 8.28 and 5.75 units respectively.
The variation(both the skein values and cost values) has the constant of 4 ie the 1st skein value × 4= the last skein value & the 1st cost value × 4=the last cost value