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## 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.

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