Can someone show by vector means that any inscribed angle in a semicircle is a right angle

Question

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.

Answer 1

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.

Answer 2

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: $$ (\vec v_2,\vec v_2)=a^2(\cos \theta -1)(\cos \theta +1)+a^2 \sin^2 \theta=a^2(\cos^2 \theta -1 + \sin^2 \theta)=0 $$

Answer 3

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.