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**1a2b3c2212****Member**- Registered: 2009-04-04
- Posts: 419

Given RD=2MD, 3RE=2RA and ED is parallel to AM,

Showing your reasons clearly,

i) Prove that ΔATE is similar to ΔART

ii) Find area of ΔATE / area of ΔART

iii) Show that ½ RT²= TM x AT

iv) Show that 3AT²=4RD²

*Last edited by 1a2b3c2212 (2010-10-07 16:27:34)*

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**Hydroex****Member**- Registered: 2010-06-14
- Posts: 13

I can can only do the first 2

:d

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,607

Ok. I'm still stuck on part (iv) so I'll throw in my attempts at parts (i) to (iii) and maybe that'll inspire someone else to complete the question.

(i) Consider triangles ATE and ART (vertices in that order)

Angle A is common to both.

Angle ATE = angle ERT (circle theorem .. angle between chord ET and tangent = angle subtended by chord ET)

=> ATE and ART are similar triangles (2 angles the same).

(ii) note: 3RE = 2RA => 2AE = RA. The three items of information all => triangle ERD is similar to triangle ARM with scale factor of enlargement 3/2.

proof

angle R is common to both

angle RED = angle RAM (AR cuts parallels in same angle)

=> RC = 2CT and perpendicular height R to AM = 3 times height C to AM

=> area ATE : area ART = 0.5 x AT x height C to AM : 0.5 x AT x height R to AM = 1 : 3

(iii) I have used a lesser known circle theorem so I'll prove it first. Refer to diagram below.

A, B, C, and D are any points on the circumference of any circle.

E is the point where AC and BD intersect.

angle ADB = angle ACB (angles subtended by the same chord)

angle DAC = angle DBC (ditto)

=> triangles AED and BEC are similar

=> AE/DE = BE/CE => AE.EC = BE.ED

Now back to the problem, where the letters are as given in the original diagram

EC.CD = RC.CT => 2/3 AT . 2/3 TM = 2/3 RT. 1/3 RT

=> AT.TM = 1/2 RT.RT

(iv) thoughts

That theorem also works if E is extrenal to the circle and when two points coincide so maybe a starting point is

MT.MT = MD.MR and AT.AT = AE.AR

Also it is possible to show that angle ART = angle TRM

which makes for a lot of similar triangles

ETC, TMD, RDC, RMT and RTE are all similar

ATE, ART, TDC, ERC and TRD are all similar

AMR and EDR are similar.

that's as far as I've got with this.

Bob

*Last edited by Bob (2010-10-08 10:59:48)*

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

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**1a2b3c2212****Member**- Registered: 2009-04-04
- Posts: 419

Hi bob bundy;

for (iii), you are proving that ECT is similar to RCD by intercepting chords theorem right?

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,607

Hi 1a2b3c2212

This is how I did that.

angle TRD = angle CET (angles subtended by the same chord TD)

angle ECT = angle RCD (vertically opposite)

To work out which angles are the same I also used corresponding angles and alternate angles, plus angle sum.

The unexpected result was that angle ART = angle TRM. I'm still wondering if the fog will clear and I'll spot an error with this one. I made an accurate copy using 'Sketchpad' and these angles were the same to 2 decimal places.

Bob

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

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**1a2b3c2212****Member**- Registered: 2009-04-04
- Posts: 419

why is there an error?

angle ATE = angle TEC (alt angle )

angle ATE = angle ART (alt segment theorem)

angle ART = angle EDT = angle CDT (angle subtended by the same chord)

hence angle TEC = angle EDT

angle TEC = angle TRD (angle subtended by the same chord)

angle EDT = angle ERT (angle subtended by the same chord)

angle TRD = angle TRM, angle ERT = angle ART

hence angle ART = angle TRM

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,607

hi

I've not explained myself well enough. I had that proof but was just surprised that those initial conditions would lead to such a result. I'm still trying to do (iv)

Bob

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

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,607

and finally ....

I've been coming back to part (iv) for a while now. I cannot get it to yield to similar triangles (of which there are many) nor the cosine rule.

In desperation, I resorted to using 'Sketchpad' to try it out. This program uses vector geometry and can be set to work to a fine degree of accuracy. The diagram below shows why I'm giving up.

The two top angles can be shown analytically to be equal. Sketchpad shows them differing by only 0.03 of a degree, which shows how accurately it can work.

The highlighted calculations show (iii), a result I stumbled upon connecting TM and RD and thirdly shows that (iv) does not seem to be correct.

1a2b3c2212

If you spot this post, please check the problem as I don't think it can be done as formulated.

Bob

*Last edited by Bob (2010-11-08 09:16:10)*

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

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