Giải thích các bước giải:

Ta có: $\Delta ABC$ cân tại $A,AH\perp BC\to H$ là trung điểm $BC$

$\to HB=HC=\dfrac12BC$

Mặt khác $AH$ vừa là đường cao vừa là phân giác góc $A$

Kẻ $HD\perp AB, HF\perp AC\to HD=HF$

Kẻ $HE\perp MN$ vì $HD\perp AB\to HD\perp MB$

Mà $MH$ là phân giác $\widehat{BMN}\to HD=HE$

$\to HD=HE=HF$

$\to HE=HF\to NH$ là phân giác $\widehat{MNC}$

$\to \widehat{MHN}=180^o-\widehat{MHB}-\widehat{NHC}$

$\to \widehat{MHN}=180^o-(180^o-\widehat{BMH}-\widehat{MBH})-(180^o-\widehat{HNC}-\widehat{HCN})$

$\to \widehat{MHN}=180^o-180^o+\widehat{BMH}+\widehat{MBH}-180^o+\widehat{HNC}+\widehat{HCN}$

$\to \widehat{MHN}=2\widehat{ABC}-180^o+\widehat{BMH}+\widehat{HNC}$

$\to \widehat{MHN}=2\widehat{ABC}-180^o+\widehat{HMN}+\widehat{HNM}$

$\to \widehat{MHN}=2\widehat{ABC}-(180^o-\widehat{HMN}-\widehat{HNM})$

$\to \widehat{MHN}=2\widehat{ABC}-\widehat{MHN}$

$\to 2\widehat{MHN}=2\widehat{ABC}$

$\to \widehat{MHN}=\widehat{ABC}=\widehat{ACB}$

$\to \widehat{MHB}=180^o-\widehat{MHN}-\widehat{NHC}$

$\to \widehat{MHB}=180^o-\widehat{ACB}-\widehat{NHC}$

$\to \widehat{MHB}=\widehat{HNC}$

Lại có: $\widehat{MBH}=\widehat{NCH}$

$\to \Delta MBH\sim\Delta HCN(g.g)$

$\to \dfrac{MB}{HC}=\dfrac{HB}{CN}$

$\to MB.NC=HB.HC=\dfrac14BC^2$

$\to MB+NC\ge 2\sqrt{MB.NC}=BC^2$

Dấu = xảy ra khi $MB=NC$