[Geometry 8] Exercises Chapter III - Similar triangles (001)

$\boxed{\text {Problem:}}$ In $\triangle ABC$, $M$ is the midpoint of the side $BC$. The bisector $ME$ of $\widehat {AMB}$ meets the side $AB$ at $E$, and the bisector $MF$ of $\widehat {AMC}$ meets the $AC$ at $F$. Prove that $\dfrac{{EA}}{{EB}} = \dfrac{{FA}}{{FC}}$. Hence $EF$ $\parallel$ $BC$.

Solution.

In $\triangle ABM$, $ME$ is the bisector of $\widehat {AMB}$.

$ \Rightarrow $ $\dfrac{{EA}}{{EB}} = \dfrac{{MA}}{{MB}}$ (angle bisector theorem).   (1)

In $\triangle$ACM, MF is the bisector of $\widehat {AMC}$.

$ \Rightarrow $ $\dfrac{{FA}}{{FC}} = \dfrac{{MA}}{{MC}}$ (angle bisector theorem).   (2)

But $MB = MC$ (M is the midpoint of BC).    (3)

(1), (2), (3) $ \Rightarrow $ $\dfrac{{EA}}{{EB}} = \dfrac{{FA}}{{FC}}$.

$ \Rightarrow $ $EF$ $\parallel$ $BC$ (converse of Thales theorem).