Giải thích các bước giải:
a.Ta có $C,F$ đối xứng qua $BE$
$\to\widehat{BFE}=\widehat{BCE}=90^o$
$\to \widehat{QFA}=90^o-\widehat{AFB}=\widehat{ABF}$
Mà $\widehat{QAF}=\widehat{BAF}=90^o$
$\to\Delta AQF\sim\Delta AFB(g.g)$
b.Gọi $BD\cap QC=G$
Ta có $F,C$ đối xứng qua $BE\to BF=BC$
Ta có $\widehat{BAF}=\widehat{BFQ}=90^o,\widehat{ABF}=\widehat{QBF}$
$\to\Delta BAF\sim\Delta BFQ(g.g)$
$\to\dfrac{BF}{BQ}=\dfrac{BA}{BF}$
$\to BF^2=BA.BQ$
$\to BC^2=CD.BQ$
$\to\dfrac{BC}{CD}=\dfrac{BQ}{BC}$
Lại có $\widehat{QBC}=\widehat{BCD}$
$\to\Delta BCQ\sim\Delta CDB(c.g.c)$
$\to \widehat{CBD}=\widehat{BQC}$
$\to \widehat{CBG}=\widehat{BQC}$
Mà $\widehat{GCB}=\widehat{QCB}$
$\to\Delta CBG\sim\Delta CQB(g.g)$
$\to \widehat{CGB}=\widehat{CBQ}=90^o$
$\to BD\perp CQ$
