1. ## geometry question

hey everybody,
im tring to solve this question for a long time without much success...
i succeeded only with trigonemtry but this is not helping because i need to solve the question with geometry only!
so here the question:

The center of the squares are Q and P
BO=OC
I Need to prove OP=OQ

Now, i found this question helpful:

on each side of Parallelogram there is a square

I need to prove that:
If the centers of the squares are connectd
a square forms in the middle
if you can prove that the square in the middle has right angles
then you can prove the first question (look at the green line its Symmetry line and also the hypotenuse)
thank you for the help
omer20

2. ## Re: geometry question

It looks as though you have made a very good start on this problem. You do not actually need to show that the "square in the middle" (which I assume is the thing with the blue diagonals) is a square. It will be sufficient to show that it is a rhombus, because the diagonal of a rhombus intersect at right angles. To do that, prove that the red triangles are congruent (two sides, included angle).

In fact, it certainly seems to be true that the "square in the middle" really is a square, but I don't offhand see how to prove that.

3. ## Re: geometry question

hey
thank you for the help
i found a way to show that this is a square:

if you can prove that the triangles are congruent (I couldnt find how to prove that one of the angles are even)
you can see that there is a right angle in the middle and then by reducing angle and adding it again you can find that the square has a right angle and then the question is solved!
but i cant find the angle so ill be glad if you write the angle you were talking about and how to prove that offcourse
thank you!!!

4. ## Re: geometry question

Originally Posted by omer20
i found a way to show that this is a square:

if you can prove that the triangles are congruent (I couldnt find how to prove that one of the angles are even) you can see that there is a right angle in the middle and then by reducing angle and adding it again you can find that the square has a right angle and then the question is solved!
Very nice!

Originally Posted by omer20
but i cant find the angle so ill be glad if you write the angle you were talking about and how to prove that offcourse
Look at the obtuse angle in the two triangles. In each case it consists of three parts. Two of these parts are 45º angles (the angle between the side of a square and the diagonal). The part in the middle, in the left red triangle, is the acute angle of the parallelogram. In the right red triangle, you can see (from the fact that all the angles at that point add up to 360º) that the part in the middle is the supplement of the obtuse angle of the parallelogram, and is therefore equal to the acute angle of the parallelogram.

5. ## Re: geometry question

thank you very much problem solved!!!

6. ## Re: geometry question

Here's another solution which i think is simpler.

$K$ is mid-point of $BP$. $L$ is mid point of $CQ$. $D$ is mid-point of $BC$.

this gives $|CP|=2|DK|, \, |BQ|=2|DL|.$.
Its easy to see that $\Delta PAC \cong \Delta BAQ \Rightarrow |BQ|=|CP| \Rightarrow |DK|=|DL|$.