Problm

I would use coordinate geometry with point B at the origin, and the line BE becomes:

[MATH]y=mx[/MATH]
Which is tangent to the circle:

[MATH](x-1)^2+\left(y-\frac{1}{2}\right)^2=\frac{1}{4}[/MATH]
Substitute for \(y\):

[MATH](x-1)^2+\left(mx-\frac{1}{2}\right)^2=\frac{1}{4}[/MATH]
Arrange in standard form:

[MATH](m^2+1)x^2-(m+2)x+1=0[/MATH]
Requiring the discriminant to be zero, we obtain (discarding the smaller root):

[MATH](m+2)^2-4(m^2+1)=0[/MATH]
[MATH]m=\frac{4}{3}[/MATH]
Hence:

[MATH]\overline{BE}=\sqrt{\left(\frac{3}{4}\right)^2+1}=\frac{5}{4}[/MATH]
 
Another method:-
diag.png

Triangles ABD and EBC are similar. Let x = length DB = length BC

Use Pythagoras on ABD:

(AD)² + (AB)² = x²

[math] \left(\frac{1}{2}\right)^2+\left(1-x\right)^2=x^2 [/math]
[math] \frac{1}{4}+1-2x+x^2=x^2 [/math]
[math] 2x=\frac{5}{4} [/math]
and since length DB=BG then [math]DG=2x=\frac{5}{4} [/math]
 
lev888 said:
We can prove that CED and CEG are similar
Click to expand...

Yes I think you're right. The quad DHCE has two right angles in it, therefore ∠HCE = 180 - ∠HDE
This means that ∠ECK = ∠HDE. This in turn means that ∠EDC=∠ECG since the red lines bisect the quads - so the triangles you indicate are similar. (There may well be a quicker way to determine this, but I can't see it at the moment!)

diag2.png

FYI: The video uses the fact that length EG=GK. Lots of ways to do this one!
 
lev888 said:
We can prove that CED and CEG are similar. This makes GE half CE.
Click to expand...

It took me a while to work out why your second sentence follows. This is because CH is half the length of HD and triangles CHD and CED are similar.

I'm going to have nightmares about similar triangles tonight ?
 
Cubist said:
It took me a while to work out why your second sentence follows. This is because CH is half the length of HD and triangles CHD and CED are similar.

I'm going to have nightmares about similar triangles tonight ?
Click to expand...
How was your nightmare?
BTW I am from India. Is there any WhatsApp or Facebook group for maths?
 
Cubist said:
It took me a while to work out why your second sentence follows. This is because CH is half the length of HD and triangles CHD and CED are similar.

I'm going to have nightmares about similar triangles tonight ?
Click to expand...
Should've explained it, but I'm on my cell phone - not the best way to enter lengthy proofs.
 
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