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Problem 14 – Solution – pg. 2/2

Posted in Solutions for Posted Problems by Lia on 05/30/2011

 Problem 14 – Solution: 1, 2

 

2. We know, from point 1. that ∠ DEF =DFE.

FD {ortho} AC and  ED {ortho} AB therefore FD {ortho} EG and  ED {ortho} FG because {{}{EG}{}} || {{}{AC}{}} and {{}{FG}{}} || {{}{AB}{}} .

That’s means that :

[3]   {Delta} EFJ = {Delta} FHE because they have two angles and one side equal:

  • JEF ={{}{HFE}{}}

  • EJF =FHE = {90^{circ}}

  • {{}{EF}{}} is hypotenuse in both triangles.

If [3] is true, ∠ HEF ={{}{EFJ}{}} therefore {Delta} EGF is an isosceles triangle. That’s means that {{}{AD}{}} is  median and height  in {Delta} EGF as well as bisector to ∠ {{}{EGF}{}}therefore {{}{G}{}} is on {{}{AD}{}}.

 

3. In the isosceles triangle {{}{EDF}{}}, KE = DK and DL = LF therefore

[4]   {{}{KL}{}} || {{}{EF}{}}.

{Delta} EFJ = {Delta} FHE(shown at [3]), therefore EH = FJ. That means that in the isosceles triangle {{}{EGF}{}},

[5]   {{}{HJ}{}} || {{}{EF}{}}.

In conclusion:

[6] {{}{KL}{}} || {{}{HJ}{}} therefore the quadrilateral {{}{JHKL}{}} ia a trapezoid.

{Delta} EKH = {Delta} FLJ because they have one angle and two sides equal:

  • HEK ={{}{LFJ}{}}(equal differences of equal angles)

  • EK = FL

  • EH = FJ

That’s means that:

[7] KH = LJ

[6] and [7] are evidence that the quadrilateral {{}{JHKL}{}} is a isosceles trapezoid.

 Problem 14 – Solution: 1, 2