a) Xét \(\Delta MAP\)và \(\Delta MDI\)có:
\(\widehat{AMP}=\widehat{DMI}\)(đối đỉnh)
\(AM=DM\)(gt)
\(\widehat{MAP}=\widehat{MDI}\) (slt do DI // AC)
suy ra: \(\Delta MAP=\Delta MDI\) (g.c.g)
\(\Rightarrow\)\(AP=DI\)
\(\Delta BPC\)có: \(DI//PC\) ; \(DB=DC\)
\(\Rightarrow\)\(IB=IP\)
\(\Rightarrow\)\(DI\)là đường trung bình \(\Delta BPC\)
\(\Rightarrow\)\(DI=\frac{1}{2}PC\)
mà \(DI=AP\) (cmt)
\(\Rightarrow\)\(AP=\frac{1}{2}PC\)
\(\Rightarrow\)\(\frac{AP}{AC}=\frac{1}{3}\) (1)
b) Kẻ \(DK//AB\) \(\left(K\in QC\right)\)
Xét \(\Delta MAQ\)và \(\Delta MDK\)có:
\(\widehat{QMA}=\widehat{KMD}\)(đối đỉnh)
\(AM=DM\)(gt)
\(\widehat{QAM}=\widehat{KDM}\) (slt do KD // AQ)
suy ra: \(\Delta MAQ=\Delta MDK\) (g.c.g)
\(\Rightarrow\)\(AQ=DK\)
\(\Delta CBQ\)có \(DK//BQ\); \(DB=DC\)
\(\Rightarrow\)\(KQ=KC\)
\(\Rightarrow\)\(DK\)là đường trung bình \(\Delta CBQ\)
\(\Rightarrow\)\(\frac{DK}{BQ}=\frac{1}{2}\)
mà \(AQ=DK\)(cmt)
\(\Rightarrow\)\(\frac{AQ}{BQ}=\frac{1}{2}\)
\(\Rightarrow\)\(\frac{AQ}{AB}=\frac{1}{3}\) (2)
Từ (1) và (2) suy ra: \(PQ//BC\)
c) \(PQ//BC\)
\(\Rightarrow\)\(\Delta MQP~\Delta MCB\)
\(\Rightarrow\)\(\frac{PQ}{BC}=\frac{MP}{BM}\)
\(\Rightarrow\)\(PQ.BM=MP.BC\) (có lẽ đề sai)
