Draw median CD of \(\Delta\)ABC. Call E is the midpoint of  CG.

From D and E draw DH and EK perpendicular with line d.

We have GQ is the midsegment of trapezium HDEK.

\(\Rightarrow GQ=\dfrac{DH+EK}{2}\Leftrightarrow2GQ=DH+EK\)

\(\Rightarrow4GQ=2DH+2EK\) (1)

Consider trapezium MABN:

DH is midsegment \(\Rightarrow2DH=AM+BN\) (2)

Similar:  \(2EK=GQ+PC\) (3)

Substitute (2) and (3) to (1): \(AM+BN+GQ+PC=4GQ\)

\(\Rightarrow AM+BN+PC=3GQ\) .

On ray BM, draw point N so AM=MN (1)

Because \(\widehat{BAM}+\widehat{MAC}=\widehat{BAC}=90^0\Rightarrow\widehat{BAM}+\widehat{MBA}=90^0\) (\(\widehat{MAC}=\widehat{MBA}\))

\(\Rightarrow\widehat{AMB}=90^0\Rightarrow\)\(\Delta AMB\) is a square triangle.

So \(\Delta MAN\) is an isosceles square triangle.

\(\Rightarrow\widehat{MAN}=\widehat{MNA}=45^0\Rightarrow\widehat{BAN}+\widehat{MAC}=45^0\)

But \(\widehat{ACM}+\widehat{MCB}=\widehat{ACB}=45^0\)

\(\Rightarrow\widehat{BAN}+\widehat{MAC}=\widehat{ACM}+\widehat{MCB}=45^0\)

\(\widehat{MAC}=\widehat{MCB}\Rightarrow\widehat{BAN}=\widehat{ACM}\)

\(\Rightarrow\Delta ABN=\Delta CAM\) (Angular angle) \(\Rightarrow BN=AM\) (2)

By (1) and (2) \(\Rightarrow AM=MN=BN\Rightarrow MB=2MA\)

So that \(MA:MB=1:2\) (*)

Consider \(\Delta MAN\): 

\(\widehat{AMN}=90^0\Rightarrow AM^2+MN^2=AN^2\Leftrightarrow2AM^2=AN^2\)(Pytagoras theorem)

\(\Rightarrow\sqrt{2.AM^2}=\sqrt{AN^2}\Rightarrow\sqrt{2}.\sqrt{AM^2}=AN\)

\(\Leftrightarrow\sqrt{2}.AM=AN\). I have \(AN=MC\) (\(\Delta ABN=\Delta CAM\))

So \(MC=\sqrt{2}.AM\) \(\Rightarrow MA:MC=1:\sqrt{2}\) (**)

From (*) and (**) \(\Rightarrow MA:MB:MC=1:2:\sqrt{2}\).

I have a picture:

From point M, draw ray MN parallel with AC. (N\(\in\)DB)

AB=AD \(\Rightarrow\)\(\Delta\)BAD is an isosceles triangle \(\Rightarrow\widehat{ABD}=\widehat{ADB}\)

But \(\widehat{MNB}=\widehat{ADB}\) (Isotopes)

So \(\widehat{ABD}=\widehat{MNB}\) or \(\widehat{MBN}=\widehat{MNB}\) 

\(\Rightarrow\Delta BMN\) is a isosceles triangle \(\Rightarrow BM=MN\).

Because BM=CD \(\Rightarrow MN=CD\).

\(MN\)//AC \(\Rightarrow\)MN//CD \(\Rightarrow\left\{{}\begin{matrix}\widehat{MNI}=\widehat{CDI}\\\widehat{NMI}=\widehat{DCI}\end{matrix}\right.\)

\(\Rightarrow\Delta MIN=\Delta CID\) (Angular angle)

Now we can prove that IM=IC (Corresponding edges)