To form the converse statement, interchange the hypothesis and the conclusion.
Converse: If D is on side CB of triangle △ ABC and CD/DB=CA/BA, then AD bisects ∠ CAB. Justification: See solution.
Practice makes perfect
To write the converse, let's identify the hypothesis and the conclusion in the Triangle-Angle-Bisector Theorem.
For a pointDon segmentCB,
ifADbisects∠ CAB
thenCD/DB=CA/AB.
In the converse we interchange the hypothesis and the conclusion.
For a pointXon segmentCB,
ifCX/XB=CA/AB
thenAXbisects∠ CAB.
This converse is true.
To justify it, let's start with the angle bisector AD and pick a point E other than D on side CB.
This point can be either to the left or to the right of D. Let's investigate the two cases separately.
If we decrease the numerator and increase the denominator of a fraction, the quotient decreases.
CEDB
⇓
CE/EBTriangle-Angle-Bisector Theorem tells us that D divides BC into two segments that are proportional to the other two sides of the triangle.
CD/DB=CA/AB
Comparing this with the inequality above gives that E does not divide BC into two segments that are proportional to the other two sides of the triangle.
CE/EB≠CA/AB
The case when E is to the right of D is similar.
E is on DB
Let's compare the ratio CEEB with the ratio CDDB.
Since E is on DB, we know that CE>CD and EB
If we increase the numerator and decrease the denominator of a fraction, the quotient increases.
CE>CDandEBCD/DB
Similarly to the case above, this gives that E does not divide BC into two segments that are proportional to the other two sides of the triangle.
CE/EB≠CA/AB
Conclusion
We know now that the only point that divides BC into two segments that are proportional to the other two sides of the triangle is the point where the angle bisector of ∠ CAB intersects side BC. This proves the converse of the Triangle-Angle-Bisector Theorem.
For a pointXon segmentCB,
ifCX/XB=CA/AB
thenAXbisects∠ CAB.
