Triangle in Circle Geometry

[Tarmac27]

Hello,

I am having trouble with the triangle inscribed in circle geometry problem. Just by looking at the LHS
I assumed that angle AEB needed to be proven to be 90 degrees in order to use pythagoras. This is where I got stuck, I just decided to see
the outcome despite not proving this angle in the end I proved what was asked but I still want to know how to prove angle AEB is 90 degrees.

Thanks.

 

[lev888]

Please post your proof and explain why you think that AEB needs to be 90 degrees. I see nothing in the problem that implies that.

 

[Dr.Peterson]

I am having trouble with the triangle inscribed in circle geometry problem. Just by looking at the LHS
I assumed that angle AEB needed to be proven to be 90 degrees in order to use pythagoras. This is where I got stuck, I just decided to see
the outcome despite not proving this angle in the end I proved what was asked but I still want to know how to prove angle AEB is 90 degrees.

You don't assume something is a right triangle in order to use the Pythagorean theorem; you use the theorem because something is a right triangle (if, in fact, it is). Do you assume a screw is a nail, so that you can use your hammer on it? No, you get a screwdriver! (Or you put a nail in somewhere else.)

Here's the figure:

Clearly AEB is not a right angle!

So you can't use Pythagoras yet. You can either add a right triangle to the figure and see if you can reach the goal using that, or look for some similar triangles (again, perhaps by adding a line or two).

You want to show that c^2 - e^2 = fg. My first thought is to construct the altitude AF = h:

 

[Tarmac27]

Hi thanks for the reply. I now realise how silly that was to assume AEB was 90 degrees. Anyway after drawing that altitude you recommened I think I have proved it:

FIrstly,

AB² = AF² + BF²

and

AE² = AF² + FE²

also,

AB² - AE² = BF² - FE²

Secondly,

BE = BF - FE

and

CE = CF + FE

Thirdly,

Trianlge AFB and Triangle AFC share two common angles and 1 common side so they are congruent. So BF = CF

Then,

BE x CE = (BF - EF) x (CF + FE)
=(BF x CF) + (BF x FE) - (EF x CF) - FE²

Recall BF = CF

BE x CE = BF² + (CF x FE) - (FE x CF) - FE²
= BF² - FE²

Hence BE x CE = AB² - AE²

Is correct? Please let me know if I have made a mistake somwhere.

Thanks

 

[Dr.Peterson]

Nicely done!

 

[Tarmac27]

Awesome,
Thanks

 

[Deleted member 4993]

Hi thanks for the reply. I now realise how silly that was to assume AEB was 90 degrees. Anyway after drawing that altitude you recommened I think I have proved it:

FIrstly,

AB² = AF² + BF²

and

AE² = AF² + FE²

also,

AB² - AE² = BF² - FE²

Secondly,

BE = BF - FE

and

CE = CF + FE

Thirdly,

Trianlge AFB and Triangle AFC share two common angles and 1 common side so they are congruent. So BF = CF

Then,

BE x CE = (BF - EF) x (CF + FE)
=(BF x CF) + (BF x FE) - (EF x CF) - FE²

Recall BF = CF

BE x CE = BF² + (CF x FE) - (FE x CF) - FE²
= BF² - FE²

Hence BE x CE = AB² - AE²

Is correct? Please let me know if I have made a mistake somwhere.

Thanks

One suggestion!

- Use ' * ' to indicate multiplication