Giải thích các bước giải:

a.Kẻ $AF\perp AN, F\in CD$

$\to \widehat{DAF}=90^o-\widehat{DAN}=\widehat{MAB}$

Mà $\widehat{ADF}=\widehat{ABM}(=90^o)$

$\to \Delta ADF\sim\Delta ABM(g.g)$

$\to \dfrac{AF}{AM}=\dfrac{AD}{AB}=k$

$\to kAM=AF$

Vì $\Delta AFN$ vuông tại $A, AD\perp NF$

$\to \dfrac1{k^2AM^2}+\dfrac1{AN^2}=\dfrac1{AF^2}+\dfrac1{AN^2}=\dfrac1{AD^2}$ không đổi

b.Ta có:

$\dfrac1{AD^2}=\dfrac1{k^2MA^2}+\dfrac1{AN^2}\ge \dfrac12(\dfrac1{kAM}+\dfrac1{AN})^2\ge \dfrac12\cdot (\dfrac4{kAM+AN})^2\ge \dfrac{8}{(kAM+AN)^2}$

$\to (kMA+AN)^2\ge 8AD^2$

$\to kAM+AN\ge 2\sqrt2AD$

Dấu = xảy ra khi $kAM=AN$

$\to \dfrac{AM}{AN}=k$

$\to \dfrac{BN}{BC}=\dfrac{AM}{AN}=k$

$\to BN=kBC$