1. ## Triangle Problem!

ABC right triangle
AC=6

We must find: AB

2. ## Re: Triangle Problem!

label intersection of BC and circle E connect DE and AE. See the solution?

3. ## Re: Triangle Problem!

please explain i dont know what you mean

4. ## Re: Triangle Problem!

Draw a line between A and the point where the circle crosses the line BC. What do you know about right angled triangles in circles?

5. ## Re: Triangle Problem!

how much did you get AB?

6. ## Re: Triangle Problem!

Meanwhile AD=AC=6 (the base of right angled triangle in a circle is on the diameter, congruent triangles. draw that line and you'll see what I mean.)
so AB=2AC=2x6=12
I cannot be bothered to write a full proof right now as this question is quite simple.'

Oh! This is interesting, it's a 30 60 90 triangle. And the triangles inside the circle are also 30 60 90. Beautiful!

7. ## Re: Triangle Problem!

thanks azzie but the answer says AB = 6*sqr(10)

8. ## Re: Triangle Problem!

Originally Posted by Azzie
Meanwhile AD=AC=6 (the base of right angled triangle in a circle is on the diameter, congruent triangles. draw that line and you'll see what I mean.)
The fact about the right angle and the diameter is correct (Thales' theorem), but it does not follow that triangles ADE and ACE are congruent and that AD = AC.

Instead, ED is both a median and an altitude in triangle ABE, so triangle ABE is...

9. ## Re: Triangle Problem!

thanks alot. i get the answer

10. ## Re: Triangle Problem!

Oops! What a foolish assumption I made, sorry about that! I obviously need to go to bed :P thanks emakarov!

11. ## Re: Triangle Problem!

BEC perpendicular to AC given AC is adiameter (subtends arc of 180
Triangle AEC 10^2 -6^2= EC^2 EC =8
ED is perpendicular bisector of AB (AD=DB given) and angle ADE =90 (subtends arc of dia)
AE=EB=10 BC = 18 18 ^2 +6^2 = AB^2 AB=18.97