Angle Inscribed in a Semicircle

Date: 11/07/2001 at 16:07:41
From: Sondra Hall
Subject: Geometry (circles)

How can I prove that any inscribed angle made in a semicircle will be 
a right angle?

Date: 11/07/2001 at 16:37:02
From: Doctor Jubal
Subject: Re: Geometry (circles)

Hi Sondra,

Thanks for writing Dr. Math.

Let's consider a circle with its center at point C.  Let's say that 
line segment PQ is a diameter of this circle, and that point R is some 
other point on the circle.  Then arc PRQ is a semicircle, and we want 
to prove that angle PRQ is a right angle.

 

Looking at this diagram, we don't know much besides the fact that P, 
Q, and R are all points on a circle with its center at C. And since 
all points on a circle are equally distant from the center, we can 
conclude that segments PC, QC, and RC are all congruent.

Now let's try to relate this to some of the angles in the diagram. If 
PC and RC are congruent, that means that PCR is an isoceles triangle, 
and the base angles CPR and CRP are congruent. For future reference, 
I'm going to call the measure of these congruent angles A. Similarly, 
since RC and QC are congruent, QCR is an isoceles triangle, and the 
base angles CQR and CRQ are congruent. Let's call the measure of these 
congruent angles B.

Since angle PRQ is the sum of angles CRP and CRQ, what we're really 
trying to prove is that A + B is 90 degrees.

The sum of the three angles in any triangle is 180 degrees.  Applying 
this to isoceles triangle PCR, we get

  2A + angle PCR = 180

and applying it to isoceles triangle QCR, we get

  2B + angle QCR = 180

and adding these two equations together, we get

  2A + 2B + angle PCR + angle QCR = 360

But PQ is a straight line segment, and it's the diameter of the 
circle, so C lies on it, and PCR and QCR are supplementary angles. So 
the sum of their measures is 180 degrees, and we now have

  2A + 2B + 180 = 360

        2A + 2B = 180

         A +  B = 90

Which is what we want to prove. Angle PRQ, which measures A + B, has 
90 degrees.  It is a right angle.

Does this help?  Write back if you'd like to talk about this some
more, or if you have any other questions.

- Doctor Jubal, The Math Forum
  http://mathforum.org/dr.math/