Giải thích các bước giải:
a.Ta có $AH\perp EF, AH$ là phân giác $\widehat{EAF}\to \Delta AEF$ cân tại A
$\to H$ là trung điểm EF $\to HE=HF$
b.Ta có:$HA\perp EF\to AH^2+EH^2=AE^2+(\dfrac{EF}{2})^2=AH^2+\dfrac{EF^2}{4}$
c.Ta có :
$\begin{split}\widehat{BME}&=180^o-\widehat{EBM}-\widehat{BEM}\\&=\widehat{ABC}-\widehat{BEM}\\&=\widehat{ABC}-\widehat{AFE}\\&=\widehat{ABC}-(\widehat{FMC}+\widehat{FCM})\\&=\widehat{ABC}-\widehat{ACB}-\widehat{BME}\end{split}$
$\to 2\widehat{BME}=\widehat{ABC}-\widehat{ACB}$
d.Kẻ $BD//AF\to \widehat{BDE}=\widehat{AFD}=\widehat{BED}\to BE=BD$
Mà $\widehat{DBM}=\widehat{MCF}(BD//CF), BM=MC, \widehat{BMD}=\widehat{FMC}$
$\to\Delta MBD=\Delta MCF(g.c.g)\to BD=CF\to BE=CF$