Circle Theorems

[royal tapir]

So I need to prove that EF = 6 or HF=4 because I know that EH and JE = 2, but I don't know how to prove that JD= 6 or HF=4, if I can prove that then I solve it using Pythagoras theorem

[keen sentinel]

Perhaps using similar triangles may help

Noting DF = 10

[cunning willow]

This is a bit more fun than I expected. There may be an easier way but here's my solution:

By Euler's theorem, since A is the incenter and B is the circumcenter, we have
$AB^2 = \sqrt{R^2-2rR}$
Where $R=5$ and $r=2$

So we have $AB = \sqrt{25-20} = \sqrt{5}$

The triangle ABC is right in C because (DF) is tangent to the circle in C.
So $BC^2+AC^2 = AB^2$
And $BC^2 = 5-4 = 1$

This means $FC = FB - BC
= 5-1=4$
Which implies $FH = 4$
And therefore $FE = 4+2 = 6$

[vapid roost]

Daddy_314

[split ingot]

Have you considered that DJ=DC and FH = FC and DC + FC = DF = twice radius = 2*5 = 10

Let’s say DJ = a, FH = b

Since DE^2 + EF^2 = DF^2, (a+2)^2 + (b+2)^2 = 10^2

So b = 10 - a, so a^2 + 4a + 4 + 12^2 -24a + a^2 = 100

So 2a^2- 20a + 12^2-100 = 0

So a^2 - 20a + 44 = 0

a = 10 - sqrt(100-44) = 10-2sqrt(14)

b = 10 - a = 2sqrt(14)

Now I am going to check if this is even possible

[cunning willow]

b is not 10-a unfortunately
Edit yes, a+b = 10

[split ingot]

Oh I see

[split ingot]

Wait why

Now it looks like it is

[split ingot]

I think this might be impossible

[cunning willow]

I think your method is correct

I just re read !

Very elegant also

Gg

We do end up with getting the solutions