Can someone show by vector means that any inscribed angle in a semicircle is a right angle Could someone explain how to prove any angle inscribed in a semicircle is a right angle using vectors. I understand that the dot product of two vectors is 0 is they are perpendicular but I don't know how to show this in a semicircle.
 A: Wlog. we can assume the semicircle to be the upper unit semicircle with centre in the plane origin $(0, 0)$. Let $(x, y)$ be a fixed point on the semicircle with $y>0$.
Consider the vector $v_1$, which is the difference between the positional vectors $(x, y)$ and $(-1, 0)$, and the vector $v_2$, which is the difference between the positional vectors $(x, y)$ and $(1, 0)$.
Now note that the angle inscribed in the semicircle is a right angle if and only if the two vectors are perpendicular. Using the scalar product, this happens precisely when $v_1\cdot v_2=0$.
So just compute the product $v_1\cdot v_2$, using that $x^2+y^2=1$ since $(x, y)$ lies on the unit circle.
A: If the semicirce has radius $a$ you can represents the two vectors as the difference between the coordinates of the points $(-a,0)$ and $(a,0)$ with respect to a generic point $(a \cos \theta, a \sin \theta)$ :
$$
\vec v_1=(a\cos \theta -a; a \sin \theta)^T \quad and \quad \vec v_2=(a\cos \theta +a, a \sin \theta)^T 
$$
so you have:
$$
(\vec v_2,\vec v_2)=a^2(\cos \theta -1)(\cos \theta +1)+a^2 \sin^2 \theta=a^2(\cos^2 \theta -1 + \sin^2 \theta)=0
$$
A: Let's consider a semicircle of radius $1$. The semicircle is in the right side of cartesian axes. So all points save for the extremes have positive $X$.
Let $P=(x,y)$ be a point in the semicircle which is not an extreme. 
$x^2+y^2=1 \implies x=+{\sqrt{1-y^2}} \implies P=(+{\sqrt{1-y^2}},y)$  
The vectors to the extremes are :
Vector to top extreme : $v_t = (0,1) - (+{\sqrt{1-y^2}},y) = ( -{\sqrt{1-y^2}},1-y)$
Vector to bottom extreme : $v_b = (0,-1) - (+{\sqrt{1-y^2}},y) = ( -{\sqrt{1-y^2}},-1-y)$  
As you said if the dot product is 0 the angle is right :
$v_t.v_b = ( -{\sqrt{1-y^2}} )^2 + (-y+1)(-y-1) = (1-y^2) + (-y)^2 - 1^2  =  0$
Rotating the semicircle and vectors construction does not change the angle, thus the property of being a right angle is general for all semicircles of radius 1.
Scaling the semicircle and vectors construction does not change the angle, thus the property of being a right angle is general for all semicircles.
A: A semicircle is one half circle.  Define 3 vectors r(sub 1), r(sub 2) and r(sub 3). From the center point of the semicircle radiate these three vectors such that say r(sub one) radiates from the center to the leftmost endpoint of the semicircle; r(sub 2) radiates from the center to the rightmost endpoint of the semicircle; and r(sub 3) radiates from the center of the circle to a locus at any point on the circle.  We then know the magnitude of r(sub 1), r(sub 2) and r(sub 3) are equal
Further two vectors, one emanating from the rightmost endpoint to the locus, and one emanating from the leftmost endpoint to the locus will form an angle; an angle which we must prove is a right angle. 
Call r(sub 1), r(sub2) and r(sub 3) vectors with origins at the center. The vectors that join the endpoints (of the semicircle) to the locus, let's call U and V (these are all vectors, please keep this in mind). Say vector U emanates from the left semicircle endpoint, and terminated at the locus. Vector V emanates from the right semicircle endpoint and terminates at the locus.
We can then redefine vector U in terms of r(sub 1), and r(sub 3) or U + r(sub  1) = - r(sub 3) and vector V can be redefined as V + r(sub 2) = - r(sub 3) {I THINK} However r(sub 1) = -r(sub 2) ...
And here, I unfortunately, lose it.  We have to show either U dot V equals zero, or the magnitude of U cross V equals one.  I suspect the cross product is an easier proof; but I keep banging my head on the dot product ... I am just a stubborn old coot.  
Then by these definitions, r(sub 1) equals -r(sub 2) (Negative r(sub 2).  we can        
