# Prove ABC,A'B'C' are congruent:D is on BC,D' is on B'C', $\angle BAD \angle CAD= \angle B'A'D' \angle C'A'D', AB=A'B', AC=A'C', AD=A'D'$

In $$\triangle ABC$$ and $$\triangle A'B'C'$$, $$D$$ is a point on line segment $$BC$$ and $$D'$$ is a point on line segment $$B'C'$$. $$\frac{\angle BAD}{\angle CAD}=\frac{\angle B'A'D'}{\angle C'A'D'}$$, $$AB=A'B'$$, $$AC=A'C'$$ and $$AD=A'D'$$. How to prove that $$\triangle ABC \cong \triangle A'B'C'$$?

If $$AD$$ and $$A'D'$$ are angle bisectors, the question is much easier: $$BD:CD=B'D':C'D'$$(angle bisector theorem), then $$\triangle ABC \cong \triangle A'B'C'$$ is proved by constructing a pair of similar triangles. But I'm stuck on the general question for days.

– Moti
Apr 23 at 20:07
• @Moti It means the ratio of the numerical value of those two angles. Apr 24 at 3:19
• Simply divide...
– Moti
Apr 24 at 4:14
• @adam-rubinson Thanks. I've edited the question just now. Apr 25 at 1:51
• You should be able to do this using the cosine rule but it gets ugly. There has to be a better way. Maybe using Heron's formula? Apr 25 at 9:38

I'm not sure how formal a proof you want, but I'll show you what I came up with.

• Draw triangle $$ABC$$.
• WLOG, let $$A'$$ have the same position as $$A$$ and $$D'$$ the same position as $$D$$ (using $$\ A'D' = AD$$).
• Try different positions of $$B'$$ and $$C'$$ using: $$1)\ A'B' = AB,\ 2)\ A'C' = AC$$ and $$3)\ \angle BAD:\angle CAD=\angle B'A'D':\angle C'A'D'.$$

For example, Suppose $$\angle B'A'D' < \angle BAD,\$$ that is, $$\exists \gamma \in\ [0,1)$$ such that $$\angle B'A'D' = \gamma (\angle BAD)$$. Then $$\angle D'A'C' = \gamma (\angle DAC)$$. But $$D'$$ is a fixed point (by assumption) and we see it is not on the line $$B'C'$$.

The green circle has centre $$A$$ and passes through $$B$$ and $$B'$$. The blue circle has centre $$A$$ and passes through $$C$$ and $$C'$$.

You can use a similar argument for $$\angle B'A'D' > \angle BAD.$$

• Is my answer faulty, or should I make it more rigorous? May 10 at 16:49