Q (1 Viewer)

[mojako]

according to my calculation, if done correctly, they are not equal.
I used the isosceles BAC and use the sine rule for triangle and that sin(90-x) = cos(x). see attachment.
theta is the angle DAC

 

[Archman]

yes they are equal: (unless specified, each triplets of letters are referring to an angle)
CBE = DAC = BAD = BED
So triangle DEB is isosceles with BD = DE
now CED = CBA and angle C is common
so triangle CDE is similar to triangle CAB
so ED/EC = BA/BC = 1
so ED = EC
recalling BD = ED = EC

To mojako: whats h?

 

[ToO LaZy ^*]

cheers!

 

[mojako]

h is length AB = BC (I think thats what it was.. I have the working at home)
I should also mention that I used BD * BC = EA * EC

 

[Archman]

I should also mention that I used BD * BC = EA * EC

no thats only true if CB = CE, which they are not in this case.

i think the identity you were looking for was CA * CE = CD * CB

 

[mojako]

yes I used that (when I did the question).
When I typed "I should also mention that I used BD * BC = EA * EC" I was thinking of the wrong thing...
and the two short vertical lines in the expression for BD are actually brackets.. print screen doesn't capture it detailed enough I guess.

My process is shown on the attachment... but it's wrong because I treat A is not being on the circle. But I'll post it anyway because I've spent considerably time re-writing it (the original working is too messy). I read your solution after re-writing mine. Bad habit