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.

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.
