Angle Inscribed in a SemicircleDate: 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/ |
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