Can someone tell me why a triangle inscribed in a semicircle is always a right triangle?

I am in Trig currently and someone told me I had to use the laws of cosine. Which I am familiar with, I just don't know how to prove it.

Thank you

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- Aug 1st 2008, 09:44 AMzodiacbravetriangle inscribed in a semicircle
Can someone tell me why a triangle inscribed in a semicircle is always a right triangle?

I am in Trig currently and someone told me I had to use the laws of cosine. Which I am familiar with, I just don't know how to prove it.

Thank you - Aug 1st 2008, 10:14 AMearboth
This is a special case of Inscribed and Central Angles in a Circle from Interactive Mathematics Miscellany and Puzzles

When you read the proof keep in mind that your central angle is 180°. - Aug 1st 2008, 03:49 PMticbol
If it should be proven by the Law of Cosines, then,

let yhe two legs of the inscribed triangle be x and y

the hypotenuse is diameter d.

we assume the angle between x and r is 90 degrees.....this angle is opposite d.

By Law of Cosines,

d^2 = x^2 +y^2 -2(x)(y)cos(90deg)

d^2 = x^2 +y^2 -2xy(0)

d^2 = x^2 +y^2 -------------this is the Pythagorean Theorem, applicable to right triangles only. So the inscribed triangle is really a right triangle, whatever the lengths of x and y are. - Aug 1st 2008, 05:26 PMQuick

It sometimes helps to see a dynamic diagram.

There should be a java applet to the right of this text (it might take a second to load)

You can move the red dots around.

Notice that angle C is always a right angle.

On a side note to others: How many knew that you could type in a java applet? I used to use it all the time, but I haven't been here in like a year. - Aug 1st 2008, 06:52 PMSoroban
Hello, zodiacbrave!

Well, here's one way . . . my way.

Quote:

Why is a triangle inscribed in a semicircle is always a right triangle?

Draw chords $\displaystyle AC = x$ and $\displaystyle BC = y.$

Let $\displaystyle \angle AOC = \theta,\;\;\angle BOC = \theta'$Code:`* * *`

* * C

* o

* r * *

*

* θ * θ' *

A o - - - - o - - - - o B

* r O r *

* *

* *

* *

* * *

Use the Law of Cosines in $\displaystyle \Delta AOC\!:\;\;x^2 \:=\:r^2 + r^2 -2(r)(r)\cos\theta $

. . We have: .$\displaystyle x^2\;=\;2r^2 - 2r^2\cos\theta$ .[1]

Use the Law of Cosines in $\displaystyle \Delta BOC\!:\;\;y^2 \;=\;r^2 + r^2 - 2(r)(r)\cos\theta'$

. . We have: .$\displaystyle y^2 \;=\;2r^2 - 2r^2\cos\theta' $ .[2]

Since $\displaystyle \theta + \theta' \:=\:180^o\quad\Rightarrow\quad \theta' \:=\:180^o - \theta$

. . $\displaystyle \cos\theta' \:=\:\cos(180^o-\theta) \:=\:-\cos\theta $

Substitute into [2]: .$\displaystyle y^2 \;=\;2r^2 - 2r^2(\text{-}\cos\theta)$

. . and we have: .$\displaystyle y^2 \;=\;2r^2 + 2r^2\cos\theta $ .[3]

Add [1] and [3]: .$\displaystyle x^2 + y^2 \;=\;(2r^2 - 2r^2\cos\theta) +(2r^2+2r^2\cos\theta) $

. . $\displaystyle x^2 +y^2 \:=\:4r^2 \quad\Rightarrow\quad x^2 + y^2 \:=\:(2r)^2$ . . . Pythagorus!

Therefore, $\displaystyle \Delta ABC$ is a right triangle with $\displaystyle \angle C = 90^o.$