(1) It is proved that the intersection e is EM vertical CD, M EN vertical AB and n respectively.

So angle EME= angle DME=90 degrees.

ENA angle = ENB angle =90 degrees.

Because CD is a high line.

So angle CDA= angle CDB=90 degrees.

So CDB angle CME angle = ENB angle =90 degrees.

So I parallel AB

Parallel optical disc

So EC/BE=MC/DM

Quadrilateral EMDN is a parallelogram.

The quadrilateral EMDN is a rectangle.

So MD=EN

Because BE=2EC

So MC/DM= 1/2.

So DM=2MC

Because angle CDA+ angle BAC+ angle ACD= 180 degrees.

So angle BAC+ angle ACD=90 degrees.

Because angle ACB= angle ACD+ angle BCD=90 degrees.

So angle BAC= angle BCD

So the tangent angle BAC = tangent angle BCD

Because tan angle BAC= 1

So tan angle BCD= 1.

In a right triangle CME, the angle CME=90 degrees.

So tan angle BCD=ME/MC= 1.

So ME=MC

So DM==EN=2MC.

EN/ME=2

Because EF is perpendicular to AE

So AEF angle =90 degrees.

Because AEF angle +EGD angle +CDB angle +EFD angle =360 degrees.

So EGD angle +EFD angle = 180 degrees.

Because angle EGD+ angle EGM= 180 degrees (equal to 180 degrees on average).

So angle EGM= angle EFD

Because angle EMD= angle ENB=90 degrees (proved)

So the triangle EMD is similar to the triangle ENF (AA).

So EG/EF=ME/NE= 1/2.

So EF=2EG.

(2) EF=EG

It is proved that in the right triangle CME, the angle CME=90 degrees.

So tan angle BCD=ME/MC

Because the tangent angle BAC=2

Angle BAC= Angle BCD (authentication)

So I /MC=2

ME=2MC

Because DM=2MC (authentication)

So DM = me

Because the quadrilateral EMDN is a rectangle (proved)

So the quadrilateral EMDN is a square

So ME=NE

Because angle EGM= angle EFD (proved)

Angle EMD= Angle ENB=90 degrees (confirmed)

So the triangle EMG congruent triangles ENF (AAS)

So EF=EG