Let $ABC$ be an isosceles triangle with $AB = AC$. Let $D, E$ and $F$ be points on line segments $BC, CA$ and $AB$, respectively, such that $BF = BE$ and such that $ED$ is the angle bisector of $\angle BEC$. Prove that $BD = EF$ if and only if $AF = EC$.
parmenides51 wrote:
Let $ABC$ be an isosceles triangle with $AB = AC$. Let $D, E$ and $F$ be points on line segments $BC, CA$ and $AB$, respectively, such that $BF = BE$ and such that $ED$ is the angle bisector of $\angle BEC$. Prove that $BD = EF$ if and only if $AF = EC$.
Let's say $\angle CAB = 2 \alpha, \angle ABE = 2 \beta$. Then $\angle BED= \frac{\angle BEC}{2} = \frac{2 \alpha + 2\beta}{2} = \alpha + \beta$. And we have that $\angle BEF = 90^o-\beta$, which means $\angle DEF = \angle DEB + \angle BEF = 90^o+\alpha$. Also we have $\angle ABC = 90^o-\alpha$. So $\angle DEF + \angle DBF = 180^o \Longrightarrow BDEF$ is cyclic.
Claim 1: If $BD=EF$, then $CE = AF$
Proof: Since $BDEF$ cyclic and $BD=EF$, we get $DE||BF$. Ss $ \alpha + \beta = \angle DEB = \angle EBF = 2 \beta \Longrightarrow \alpha = \beta \Longrightarrow AE=BE=BF$. From $ABC$ is isosceles we get $AC = AB \Longrightarrow AE + CE = AF + FB \Longrightarrow \boxed{CE=AF}$.
Claim 1: If $CE=AF$, then $BD = EF$
Proof: Since $ABC$ is isosceles $AE + CE = AF + FB \Longrightarrow AE=BF=BE$ So again $\alpha = \beta$, so $DE||BF$. Which means $BDEF$ is isosceles trapezoid so $\boxed{BD=EF}$