Ta có: \(\hat{MHD}+\hat{MHE}=\hat{DHE}=90^0\)
\(\hat{MHE}+\hat{PHE}=\hat{MHP}=90^0\)
Do đó: \(\hat{MHD}=\hat{PHE}\)
Xét ΔMHD và ΔPHE có
\(\hat{MHD}=\hat{PHE}\)
\(\hat{HMD}=\hat{HPE}\)
Do đó: ΔMHD~ΔPHE
=>\(\frac{MH}{PH}=\frac{MD}{PE}\)
=>\(MH\cdot EP=MD\cdot HP\)

