Giải thích các bước giải:

a.Xét $\Delta AMB,\Delta AMD$ có:
Chung $AM$

$\widehat{MAB}=\widehat{MAD}$

$AB=AD$

$\to \Delta AMB=\Delta AMD(c.g.c)$

b.Xét $\Delta AIB,\Delta AID$ có:

Chung $AI$

$\widehat{IAB}=\widehat{IAD}$

$AB=AD$

$\to \Delta IAB=\Delta IAD(c.g.c)$

$\to \widehat{AIB}=\widehat{AID}$

Mà $\widehat{AIB}+\widehat{AID}=180^o$

$\to \widehat{AIB}=\widehat{AID}=90^o$

$\to AI\perp DB$

c.Từ a $\to MB=MD, \widehat{ABM}=\widehat{ADM}$

$\to \widehat{MBH}=180^o-\widehat{ABM}=180^o-\widehat{ADM}=\widehat{MDC}$

Xét $\Delta MBH,\Delta MDC$ có:

$\widehat{MBH}=\widehat{MDC}$

$MB=MD$

$\widehat{BMH}=\widehat{DMC}$

$\to \Delta MBH=\Delta MDC(g.c.g)$

$\to MH=MC, BH=DC$

$\to AH=AB+BH=AD+DC=AC$

Do $P$ là trung điểm $CH$

$\to PH=PC$

$\to A, M, P\in$ trung trực $CH$

$\to A, M, P$ thẳng hàng