# A geometry problem with a circle and a bisected radius

## 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.

Last edited:

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BvU
Homework Helper
2019 Award
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 ...

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
Homework Helper
2019 Award
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

berkeman
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
Mentor
But the problem statement says:
In triangle ABC, BAC=90°. AD is an altitude of the triangle ABC.