Giải thích các bước giải:

a.Gọi $OE\perp CD\rightarrow E$  là trung điểm CD

$\rightarrow EC=ED\rightarrow EC+CM=ED+DN\rightarrow EM=EN$

Mà $OE//AM$

$\rightarrow AM//OE//BN\rightarrow BN\perp CD $

b.Kẻ $IH\perp AB, IK\perp CD$

Vì $\widehat{IHA}=\widehat{CMA}=90^o$

     $\widehat{IAH}=\widehat{CAM}(+\widehat{ACM}=90^o)$

$\to\Delta IHA\sim\Delta CMA(g.g)$

$\to \dfrac{S_{IHA}}{S_{CMA}}=\dfrac{IH^2}{CM^2}\to S_{IHA}=\dfrac{IH^2}{CM^2}\cdot S_{CMA}$

Tương tự chứng minh được: $S_{IHB}=\dfrac{IH^2}{DN^2}\cdot S_{BDN}$

$\to S_{IHA}+S_{IHB}=\dfrac{IH^2}{CM^2}\cdot S_{CMA}+\dfrac{IH^2}{DN^2}\cdot S_{BDN}$

$\to S_{IHA}+S_{IHB}=\dfrac{IH^2}{DN^2}\cdot S_{CMA}+\dfrac{IH^2}{DN^2}\cdot S_{BDN}$

$\to S_{IHA}+S_{IHB}=\dfrac{IH^2}{DN^2}\cdot (S_{CMA}+S_{BDN})$

$\to S_{AIB}=\dfrac{IH^2}{DN^2}\cdot (S_{CMA}+S_{BDN})(1)$

Hoàn toàn tương tự chứng minh được:

$S_{CID}=\dfrac{IK^2}{DN^2}(S_{CM}+S_{BDN})(2)$

Vì $\Delta IDK$ vuông ở $K\to IK^2=ID^2\sin^2\widehat{IDK}$

      $\Delta BDN$ vuông ở $N\to DN^2=BD^2\sin^2\widehat{NBD}$

      $\Delta IHB$ vuông ở $H\to IH^2=IB^2\sin^2\widehat{IHB}$

Mà $\widehat{IDK}=\widehat{NBD}=\widehat{IBH}\to IH^2=DN^2+IK^2(3)$

Từ $(1), (3)$

$\to S_{AIB}=\dfrac{IK^2+DN^2}{DN^2}(S_{CMA}+S_{BDN})$

$\to S_{AIB}=(\dfrac{IK^2}{DN^2}+1)(S_{CMA}+S_{BDN})$

$\to S_{AIB}=(1+\dfrac{IK^2}{DN^2})(S_{CMA}+S_{BDN})$

$\to S_{AIB}=S_{CMA}+S_{BDN}+\dfrac{IK^2}{DN^2}(S_{CMA}+S_{DBN})$

$\to S_{AIB}=S_{CMA}+S_{BDN}++S_{CID}$ vì $(2)$

$\to đpcm$