Triangle Problem

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HellloooGuyyys

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I found this question randomly and was not able to solve it. It would be great if you could help me.

Let ∆ABO be a right triangle with the right angle at O. Let AO=a and BO=b and let h be the perpendicular bisector of AB. Be C the intersection of h with a. Show for a>b that |BC|= (a^2+b^2)/2a
 
HellloooGuyyys said:
I found this question randomly and was not able to solve it. It would be great if you could help me.

Let ∆ABO be a right triangle with the right angle at O. Let AO=a and BO=b and let h be the perpendicular bisector of AB. Be C the intersection of h with a. Show for a>b that |BC|= (a^2+b^2)/2a
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Please show us what you have tried and exactly where you are stuck.

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Please share your work/thoughts about this problem.
 
I am currently grade 10 and found this following problem where I'm now stuck on.
"Let ABO be a right triangle with the right angle at the point O and AO=a and BO=b. The perpendicular bisector of AB intersects AO in the point C. We consider a>b for the following problems.
a) Show that |BC|=(a^2+b^2)/(2a)
b) Find a pair (a,b) such that |BC| is an integer.
c) Find a pair (a,b) such that the triangle ABC has integer sides and altitudes."

a) I call S the point of the perpendicular bisector on AB. I showed that the |AC|=|BC| due to the SAS congruence of the triangles ASC and CSB. I found that AS=sqrt(a^2+b^2)/2 but I am not sure how to continue. I think I have to find a formular for SC, but I wasn't yet able to find one.
b) was pretty easy for me. I set a=4k and b=4l=a/2, k>l, 2|k and then showed that 2a divides a^2+a^2/4=
a^2+a*4k/4=a(a+k)
c) I have unfortunately no clue where to start. All I know is that |SC|, |SB| and |BC| have to be a pythagorian triple.

Thank you for your help :)
 
Can you please post a labeled triangle so we can see where you are at?
 
I don't no how a can find a formula for s0 and I don't even know if I need to, in order to prove the formula of r1
IMG_20200829_125953__01.jpg
 
Now solve for so2 in terms of r1 and s1.

The work you have done so far is nice.
 
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