In the Diagram,ABC is a right angle, AC=1 and angle BAC = alpha.

Points D and E are on the line AC such that CD=CE=CB.

a)Express the lengths AB and BC in terms of alpha.

b)Hence prove that

1/AD + 1/AB = 2/(AB)^2

Well, I have a problem in "b" part.

I have no idea how to start this Question .Pythagoras theorem,Similarity or area method used but no friutful result. Please guide.

Points D and E are on the line AC such that CD=CE=CB.

a)Express the lengths AB and BC in terms of alpha.

b)Hence prove that

1/AD + 1/AB = 2/(AB)^2

Well, I have a problem in "b" part.

I have no idea how to start this Question .Pythagoras theorem,Similarity or area method used but no friutful result. Please guide.

Attachment not found

Original post by alevelmath1

In the Diagram,ABC is a right angle, AC=1 and angle BAC = alpha.

Points D and E are on the line AC such that CD=CE=CB.

a)Express the lengths AB and BC in terms of alpha.

b)Hence prove that

1/AD + 1/AB = 2/(AB)^2

Well, I have a problem in "b" part.

I have no idea how to start this Question .Pythagoras theorem,Similarity or area method used but no friutful result. Please guide.

Points D and E are on the line AC such that CD=CE=CB.

a)Express the lengths AB and BC in terms of alpha.

b)Hence prove that

1/AD + 1/AB = 2/(AB)^2

Well, I have a problem in "b" part.

I have no idea how to start this Question .Pythagoras theorem,Similarity or area method used but no friutful result. Please guide.

Attachment not found

Can you try uploading a pic of the question agiain?

The hence part suggests going down a trig identity route, but Im not sure your posted question is correct.

(edited 1 year ago)

I have send the picture again .Please check.

AB = Cosalpha ,BC=Sinalpha. I even tried to find alpha angle which give me solution like alpha =90, which is not possible.So I have a doubt in this questuion or may I have missing something. I also figured out that triangle EBD is right angle at B.So still confusion is remained.

Original post by alevelmath1

AB = Cosalpha ,BC=Sinalpha. I even tried to find alpha angle which give me solution like alpha =90, which is not possible.So I have a doubt in this questuion or may I have missing something. I also figured out that triangle EBD is right angle at B.So still confusion is remained.

You can't find the actual angle alpha, rather its an identity you have to show all apha (all relevant side length). So what is the identity you want to show in terms of trig funtions of alpha?

could you figure out or show me any relevant solved exampme ? As uptil now there is no clear cut answer for our discussion.

Original post by alevelmath1

could you figure out or show me any relevant solved exampme ? As uptil now there is no clear cut answer for our discussion.

You dont seem to be taking the hint. What is AD and AE in terms of alpha (theyre simple expressions)?

Then rewrite the desired side length identity in terms of a trig identity. Its then a couple of simple lines to show that its true.

(edited 1 year ago)

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