# Let $ABC$ be a triangle with $AC>AB$, a median is drawn from the vertex $A$ intersecting the side $BC$ at $D$. Prove that $\angle BAD>\angle CAD$.

Question : Let $$ABC$$ be a triangle with $$AC>AB$$, a median is drawn from the vertex $$A$$ intersecting the side $$BC$$ at $$D$$. Prove that $$\measuredangle BAD>\measuredangle CAD$$.
[This was in My Nephew's Textbook who hasn't studied Trigonometry Yet]

I solved it using the Law of Sines, but I couldn't prove it without trigonometry. Can someone help?

You draw a parallelogram $ABEC$ with $D$ is midpoint of $AE$. Then you will have $\angle CAD= \angle DEB$ and $AC=BE$. Triangle $ABE$ has $AB<BE$ (because $BE=AC$) so $\angle DAB > \angle DEB$ but $\angle DEB= \angle CAD$ so done.