a) Ta có:
$\widehat{AACM} = \widehat{AAB} + \widehat{BAC} + \widehat{ACM} = \widehat{AAB} + \widehat{BAC} + \widehat{ABC} = 180^\circ - \widehat{CAB} = \widehat{AAQ} + \widehat{QAC} + \widehat{ACM} = \widehat{AAQP}$
Vậy $\widehat{AACM} = \widehat{AAQP}$, suy ra $AACM = AAQP$.

b) Ta có:
$\widehat{AHM} = \widehat{AHC} + \widehat{CHM} = \widehat{AEC} + \widehat{CBM} = \widehat{AEC} + \widehat{ACB} = 180^\circ - \widehat{EAC} = \widehat{EAP} = \widehat{QAP}$
Vậy $\widehat{AHM} = \widehat{QAP}$, suy ra $AH \parallel PQ$.
Do đó, ta có $AHIP$ là hình bình hành, nên $AH = IP$.
Tương tự, ta có $AH = IQ$.
Vậy $AH = IP = IQ$.

c) Ta có:
$\widehat{BQH} = \widehat{BCH} = \widehat{BAC} = \widehat{BAM}$
Vậy $BQ \parallel AM$.
Do đó, ta có $\triangle BQH \sim \triangle AMH$.
Từ đó, ta có $\frac{BQ}{AM} = \frac{BH}{AH}$.
Vì $BH = 2HM$, nên $\frac{BQ}{AM} = \frac{2HM}{AH}$.
Mà $AH = 2IP = 2AM$, nên $\frac{BQ}{AM} = \frac{2HM}{2AM}$.
Suy ra $BQ = 2AM$.