Geometry problem! Can anyone...

[SilverSprite]

Can anyone find a geometrical solution to this problem. I dont mean trig. or anything else of the sort. Good luck.

If, in triangle ABC, median AM is such that angle BAC is divided in ratio 1:2, and AM is extended through M to D so that DBA is a right angle, prove that AC=AD/2:

http://www.vbforums.com/attachment.php?s=&postid=1266509

[opus]

Here you go,
draw a circle with a radius of 1
The centre is point A
draw a ghorizontol line thru A
Draw anothe line thru A at an angle BAC (from your drawing), extend the line to a length of 0.5 from A, the end is point C
Draw another line thru A at an angle of 1/3 of BAC, extend that line to a length of 1 (to the perimeter of the circle. This point is D.
From D draw a vertical line the the horizontal line, they meet at point B.
You will have an Triangle ABC, and a line AD with length of AC=0.5*AD, and angle BAC is 3* the angle of BAD.

[bugzpodder]

hmm this doesn't look too much like a proof for me. besides I can't really follow it (lost at line 5)

[Kalkewl8ter]

opus,

Could you explain a bit more clearly? Perhaps with a drawing? I'm completely lost...

[opus]

He didn't ask for a proff, just for a solution!
I hope the drawing helps

[bugzpodder]

he did
prove that AC=AD/2:

[Kalkewl8ter]

Prove that AM is the median of triangle ABC, and you'll have your proof...
Although this looks rather fishy...Since AC is the radius of the smaller circle, and BC is tangent to it, would ACB not have to be a right angle? Unless you didn't mean for BC to be tangent to the circle...?

[opus]

I didn't do a proof, just a solution of how to construct it!
The circles HAVE to be radius 1 and 0,5; that makes AC=AD/2
And I did use to angles , one was the angle formed by BAC and the other 1/3 of this. This way the later will dived the first angle by a ratio of 1:2, what other proof do you need?
And for the line BC, it goes vertical from the perimeter of the small circle, that makes it a tangent.
I just tried to give a method of whow to do it, without mentioning any trigometrie, since SilverSprite said :I dont mean trig

[Kalkewl8ter]

Come on, people! Are you just going to give up on this problem? Have you no dignity whatsoever, or are you just lazy? Do none of you like geometry? This thread has been up forever and no one has solved the problem yet! I can't do it, as geometry is a very weak area for me, but there must be at least one person on this forum with geometrical insight!

[bugzpodder]

Let <BAM=x
then MAC=2x
choose point P on AD so that AM=MP.

since BM=MC, ACPB is a parallelgram (diagonals bisect each other). Thus BP=AC.
Let T be the midpoint of AD making BT the median of right triangle ABD. It follows that BT=AD/2, or BT=AT (measure of medians on the hypotneuse is half the hypotneuse length). and consequently, <TBA=x. <BTP is an exterior angle of isosceles triangle BTA. Therefore, <BTP=2x. However, since BP is parallel to AC, <CAP=<BPA=2x. Thus TBP is isosceles with BT=BP.
Since BT=AD/2 and BT=BP=AC, AC=AD/2

[SilverSprite]

Unfair, you looked up the solution!

[bugzpodder]

so? it isn't for you its for Kalk!

[SilverSprite]

:P:P I like your trig solution BTW.

[Kalkewl8ter]

Nice solution, even if you looked it up, bugz!

[SilverSprite]

Hey bugz, supposedly Yang found an easy proof. One that takes like 2 lines. Talk to her about it.

[Kalkewl8ter]

More like 8 lines.:rolleyes:

[SilverSprite]

Do tell Kalk!

[bugzpodder]

I'm all ears!

[Kalkewl8ter]

I don't remember it, but it has something to do with finding angles. Ask Yang.