Math Is Fun Forum

Monox D. I-Fly

Member

From: Indonesia

Registered: 2015-12-02

Posts: 2,000

[ASK] Question About Circle and Parallelogram

A parallelogram ABCD has angle A = angle C = 45°. Circle K with the center C intercept the parallelogram through B and D.  AD is extended so that it intercepts the circle at E and BE intercepts CD at H. The ratio of the area of triangle BCH and triangle EHD is ....

Here I got that triangle BCH and triangle EHD is similar with angle BCH = angle HDE = 45°, angle CHB = angle DHE = 112.5°, and angle CBH = angle DEH = 22.5°. The area of triangle BCH is ½ × CH × h, where h is the parallelogram's height. The area of triangle EHD is ½ × (r - CH) × r. I stuck at the ratio is (CH × h) : ((r - CH) × r). How do I simplify that? Problem is, I don't know CD's length, so I can't approximate the ratio between CH and DH.

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Actually I never watch Star Wars and not interested in it anyway, but I choose a Yoda card as my avatar in honor of our great friend bobbym who has passed away.
May his adventurous soul rest in peace at heaven.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,619

Re: [ASK] Question About Circle and Parallelogram

hi Monox D. I-Fly

Does this look right for the problem?

The parallelogram must be a rhombus  as opposite sides are equal for any parallelogram and CD = CB =radius = R for the circle.

I have drawn AC extended to cut the circle at F and G.

There's a circle theorem that should be useful here.  But it's 50 years since I last used it so the exact form escapes me at present.

Choices: (1) Go up in the attic and find an old maths book.  (2) Search through Euclid until I find it.  Or (3) Try to re-prove it from scratch.  Pride requires that I try (3) first.

Bob

ps.  You wanted CD.  If the shape is a rhombus then AB = BC = CD = DA.

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Bob

Administrator

Registered: 2010-06-20

Posts: 10,619

Re: [ASK] Question About Circle and Parallelogram

OK, I have it.  Draw any circle and mark a point T outside the circle.  Draw two lines from T; the first TPQ where P and Q are points on the circumference, and similarly, TRS.

Angles RSP and RQP are equal as the they subtended by the same chord RP so triangles TPS and TRQ are similar.  Hence  PT . TQ = RT . TS

Applying this to the problem: AD . AE = AF . AG  therefore AE = AF . AG / AD

And so DE = AF . AG / AD - AD.

Let BC = R, then AC = 2R cos(22.5) and so AF = R(2 . cos 22.5 - 1)

This allows you to get DE in terms of R and the cosine and hence DE / BC.

The area ratio is the square of this as the triangles are similar.

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Monox D. I-Fly

Member

From: Indonesia

Registered: 2015-12-02

Posts: 2,000

Re: [ASK] Question About Circle and Parallelogram

hi Monox D. I-Fly

Does this look right for the problem?

https://i.imgur.com/ilpQUFj.gif

No, ABCD is named counter clockwise from the bottom left.

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Actually I never watch Star Wars and not interested in it anyway, but I choose a Yoda card as my avatar in honor of our great friend bobbym who has passed away.
May his adventurous soul rest in peace at heaven.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,619

Re: [ASK] Question About Circle and Parallelogram

OK, try this:

Looks to me like the same problem as seen in a mirror.

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Monox D. I-Fly

Member

From: Indonesia

Registered: 2015-12-02

Posts: 2,000

Re: [ASK] Question About Circle and Parallelogram

Well, anyway, it's solved. Thanks for your help.

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Actually I never watch Star Wars and not interested in it anyway, but I choose a Yoda card as my avatar in honor of our great friend bobbym who has passed away.
May his adventurous soul rest in peace at heaven.