question on geometry

[Martoman]

Now I get to ask a question

Q8 in short answer from Ch:1 toolbox.

For the lazy ones I understand your pain so all you have to do is click this link to see the question:

http://img130.imageshack.us/f/capturefj.png/

I can get the answer. I want a more elegant way rather than the hacky crap i've been doing.

Thanks y'all 

[theuncle]

Not sure if you would call this hacky, but I'll give it a go!

Part a uses the alternate segment rule on page 21, as well as straight lines and triangles adding to 180.
Given the answer is <BCD=x, ACD is an isosceles triangle.

ABD is also an isosceles triangle given that AB=BD
an application of the sin rule gives two equations:


The rest is algebra, equating both to and solving the remaining equation for y!

[Martoman]

yeah that was the *hacky* maths I was referring to, that was the route I took.

Thanks a lot for answering though, no one else seemed to love me enough 

[98.40_for_sure]

Wouldn't have a clue! geometry is the topic i can't do

[Martoman]

Wouldn't have a clue! geometry is the topic i can't do

He rhymes without meaning to! (lol c what i did thar?) 

I love geometry lots. its pretttttyyyy  :smitten:

[Ahmad]

This follows directly from ABD and ACD being similar triangles, AB/AD = AD/AC.

To show the similarity it suffices to show ABD = ADC since they share the common angle at A. Well, COD = 2 * CBD = 2 * (180 - ABD) and OCD is isosceles giving OCD = 1/2 (180 - 2(180 - ABD)) = ABD - 90. This gives ADC = 90 + ODC = ABD as required.

[98.40_for_sure]

Wouldn't have a clue! geometry is the topic i can't do

He rhymes without meaning to! (lol c what i did thar?) 

I love geometry lots. its pretttttyyyy  :smitten:

Hahah Martoman, you are too good at picking up the nitty gritty... boo!