Let's try to draw different triangles â–ł ABC, where the angle bisector AD divides side BC into segments with lengths BD=6cm and CD=9cm.
We do not know the lengths of AB and AC, but the Triangle-Angle-Bisector Theorem tells us the ratio of these two sides.
BD/CD=AB/AC
Since it is given that BD=6 and CD=9, we can use multiples of these to preserve the ratio.
AB=6xandAC=9xfor somex
The question is can we choose x to be any positive number, or is there a restriction?
Let's see what we get when we check the Triangle Inequality Theorem.
AB+AC&>BC
AB+BC&>AC
BC+AC&>AB
Let's check the three inequalities one at a time.
AB+AC>BC
Let's substitute AB=6x, AC=9x, and BC=6+9=15 and solve the inequality.
This is true for all positive x, so BC+AC>AB is always true.
Conclusion
Putting all the results together, we got the possibilities for the lengths of the other two sides of the triangle.
AB=6xandAC=9x for some1
Extra
What is the locus of the third vertex?
It is interesting to note that the locus of vertex A as we look at all possible triangles is a circle. Can you prove this? Can you find the radius of this circle?
