[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 // 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) $\Rightarrow$ $\dfrac{{EA}}{{EB}} = \dfrac{{FA}}{{FC}}$

$\Rightarrow$ DE // BC (converse of Thales theorem)