MHB How do I calculate the side of a rhombus using the bisector of an angle theorem?

[Yankel]

Hello all,

I have this question I struggle with...

View attachment 7836

EDFB is a parallelogram. It is known that AB/BC = AD/DC.

1) Prove that the parallelogram is a rhombus.

2) It is given that: AB=9, AC=10, BC=AD. Calculate the side of the rhombus.

I think I solved the first part. There is a theorem called the "bisector of an angle theorem" according to which if AB/BC = AD/DC then the line BD is a bisector of an angle of the angle B and then a parallelogram in which the diagonal is a bisector of an angle is a rhombus. Am I correct ?

I have a problem with the second part. I can't figure out how to solve it. The answer should be 3.6. I have tried the intercept theorem (or Thales' theorem), but couldn't figure it out.

Can you kindly assist to in the second part of the question ?

Thank you in advance !

 

[Opalg]

Hello all,

I have this question I struggle with...
EDFB is a parallelogram. It is known that AB/BC = AD/DC.

1) Prove that the parallelogram is a rhombus.

2) It is given that: AB=9, AC=10, BC=AD. Calculate the side of the rhombus.

I think I solved the first part. There is a theorem called the "bisector of an angle theorem" according to which if AB/BC = AD/DC then the line BD is a bisector of an angle of the angle B and then a parallelogram in which the diagonal is a bisector of an angle is a rhombus. Am I correct ?

Yes.

I have a problem with the second part. I can't figure out how to solve it. The answer should be 3.6. I have tried the intercept theorem (or Thales' theorem), but couldn't figure it out.

Can you kindly assist to in the second part of the question ?

If BC = AD = x, then the equation AB/BC = AD/DC becomes 9/x = x/(10-x), a quadratic for x with a unique positive solution.

From the similar triangles AED and ABC you can then calculate that ED/x = x/10, which gives ED = 3.6.