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
\[\frac{DA}{DC} = \frac{DC}{DB}\]
therefore
\[ DC^{2} = DA . DB = ab\]
and thus \[DC = \sqrt{ab}\]
if r is the radius of the semicircle , then \[DC \leq r\]
we know \[ r = \frac{diameter}{2} = \frac{a + b}{2}\]
Thus
\[\boxed{ \frac{a + b}{2} \geq \sqrt{ab}}\]
If \[DC = r\] that is if the Point D comes on the center of the circle then \[a = b\]
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Top NewestGeometry proves AM-GM. That's the beauty of mathematics. Reshared! – Sanjeet Raria · 2 years, 11 months ago
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Great @megh choksi Keep it up bro – Deepanshu Gupta · 2 years, 11 months ago
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How to mention someone on brilliant? – U Z · 2 years, 11 months ago
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Just write @Name which You want to post – Deepanshu Gupta · 2 years, 11 months ago
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Geometry can also prove RMS, AM, GM, HM too. Look at here. – Samuraiwarm Tsunayoshi · 2 years, 11 months ago
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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? – Ahmed Yaseen · 2 years, 11 months ago
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