A geometry problem with a circle and a bisected radius

[Akash47]

Homework Statement

In triangle ABC, BAC=90°. AD is an altitude of the triangle ABC. A circle is drawn which center is A and radius is AD (I have got the problem from a book.But there's a printing mistake and the problem states that the radius is just 'A'!! But 'A' is just a point and so it can't be radius.So I have just guessed it should be 'AD' but I'm not sure about it.).The circle intersects triangle ABC at U and V. UV meets with AD at E.Prove that AE=DE.The picture is not drawn to scale.

Relevant Equations

No equation is required.

I have tried a lot by angle chasing e.g. let ∠ABC=x° then ∠ACB=90°-x°. As AU=AV=radius of circle so ∠AUV=∠AVU=45°. I've connected U,D and V,D. Then ∠UDV=135° etc. But I haven't found any way to get near of proving AE=DE. I have also tried to prove 'the area of triangle AEU= area of triangle DEU'.But I've also failed this time.So please tell me what to do next.

 

[BvU]

Hi,

Here ∠AUV=∠AVU=45°

That is not in the problem statement. Is it a given, or have you made an assumption ?
Whatever, even with 45° I don't agree that AE=DE can be right ...

 

[Akash47]

Hi,
That is not in the problem statement. Is it a given, or have you made an assumption ?
Whatever, even with 45° I don't agree that AE=DE can be right ...

No,it's not the statement. Please now look in the problem. I've said that I've just guessed that The radius should be AD,but I'm not sure.

 

[BvU]

Counter example: let ∠DAC indeed be 45°, and the radius be $r$. Then AE = ${1\over 2} r \sqrt 2\$ so clearly AE $\ne$ DE

 

[Akash47]

Your example is not reasonable.I have just guessed that 'AD ' is radius.Then clearly ∠ AUV=∠AVU=45°.But how can you say that ∠DAC=45°?This is totally unreasonable. Have I ever said that ∠ BCA=45°?Please have a good look on the problem.

 

[berkeman]

Your example is not reasonable.I have just guessed that 'AD ' is radius.

But the problem statement says:

In triangle ABC, BAC=90°. AD is an altitude of the triangle ABC.

which makes AD the radius of the circle.

 

[Mark44]

Also, since ∠BAC = 90°, and since U and V are points on the circle, AU and AV are radii of the circle. From this we can infer that ∠AUV = ∠AVU = 45°.

Your example is not reasonable.I have just guessed that 'AD ' is radius.Then clearly ∠ AUV=∠AVU=45°.But how can you say that ∠DAC=45°?This is totally unreasonable.

@BvU's example is reasonable. He's not saying that ∠DAC=45°; he's using that as a counterexample to what you're supposed to prove. His single counterexample shows that you cannot prove that AE = ED.