Lets us consider a semicircle with diameter AB. If C is any point on the semicircle, then ABC forms a right angled triangle with right angle at C .
Let CD be the perpendicular to AB dropped from C . Let AD= a and DB=b
The trianges DAC and DCB are similar , hence
therefore
and thus
if r is the radius of the semicircle , then
we know
Thus
If that is if the Point D comes on the center of the circle then
Please reshare if you liked it
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
Sort by:
Top NewestGeometry proves AM-GM. That's the beauty of mathematics. Reshared!
Log in to reply
Great @megh choksi Keep it up bro
Log in to reply
How to mention someone on brilliant?
Log in to reply
Just write @Name which You want to post
Log in to reply
Geometry can also prove RMS, AM, GM, HM too. Look at here.
Log in to reply
the only condition when DC = r will be when D is the centre of circle. And taking D as the centre of circle will it self make AD = a = radius = b = DB, why do we even need such a long calculation?
Log in to reply