Giải thích các bước giải:

a.Xét $\Delta DEK,\Delta DFH$ có:

$DK=DH$

Chung $\hat D$

$DE=DF$

$\to \Delta DEK=\Delta DFH(c.g.c)$

$\to EK=HF$

b.Từ a $\to\widehat{DEK}=\widehat{DFH},\widehat{DKE}=\widehat{DHF}$

$\to\widehat{OEH}=\widehat{OFK},\widehat{OHE}=180^o-\widehat{DHF}=180^o-\widehat{DKE}=\widehat{OKF}$

Xét $\Delta HOE,\Delta KFO$ có:

$\widehat{OHE}=\widehat{OKF}$

$HE=DE-DH=DF-DK=KF$

$\widehat{OEH}=\widehat{OFK}$

$\to \Delta OHE=\Delta OKF(g.c.g)$

c.Từ b $\to OE=OF$

Mà $DE=DF$

$\to D, O\in$ trung trực $EF$

$\to DO$ là trung trực $EF$

$\to DO\perp EF$