Plane Geo (1 Viewer)

[khorne]

Hey, I have a question (Further 2U) from plane geo. The diagram is attached.

The question says: AB=BC=CD, BCFE is a parallelogram with BE = 2BC. Prove ED is perpendicular to AF.

I did it, but it's very untidy. I proved congruency for triangles on the CF side, and thus proved they were isosceles, etc.

However, can anyone think of a more concise proof?

 

[Drongoski]

Hey, I have a question (Further 2U) from plane geo. The diagram is attached.

The question says: AB=BC=CD, BCFE is a parallelogram with BE = 2BC. Prove ED is perpendicular to AF.

I did it, but it's very untidy. I proved congruency for triangles on the CF side, and thus proved they were isosceles, etc.

However, can anyone think of a more concise proof?

Let AF intersect BE at P, DE intersect CF at Q

Then it can be shown:

Triangle EPF congruent to triangle BPA so that BP = PE so that P is mid-pt of BE
Similarly Q is mid-pt of CF

.: EP = EF = QF

Now triangle PEQ and triangle FQE are also congruent so that PQ = EF

.: PE = EF = FQ = PQ so that PQFE is a rhombus

Since PF and EQ are diagonals of the rhombus, they intersect at right angles.

Hence ED perp AF

This proof may be worse !

 

[khorne]

That's exactly what I did =/ I thought there was an easier way, but there probably isn't

 

[Drongoski]

That's exactly what I did =/ I thought there was an easier way, but there probably isn't

There are various ways, some possibly neater. I just tried the 1st one that came to mind.