Prove that the angle inscribed in a semi-circle is a right angle using Vector Methods. (1 Viewer)

[RohitShubesh21]

I Am Interested in if there is other ways it is done

 

[tickboom]

Is this essentially Thale's theorem? ... here's how I proved it without vectors ...

 

[A1La5]

Attempt this:

 

[ExtremelyBoredUser]

Attempt this:

View attachment 33590

Thats the other way I thought of. I prefer the conventional way though but this is neat anyways. Not sure if there's any other way other than those 2 really.

 

[A1La5]

Thats the other way I thought of. I prefer the conventional way though but this is neat anyways. Not sure if there's any other way other than those 2 really. I assume the first method that tickboom showed was the "vector" approach OP was looking for but its well known so maybe this is.

The derivation shown in the video uses Euclidean geometry instead of vectors. I also prefer his approach, but given how OP wanted a vector based approach the question I posted should be content.

 

[ExtremelyBoredUser]

I Am Interested in if there is other ways it is done

iirc there's a way to show this through showing that the vectors that inscribe the semi-circle are perpendicular through the normal equation. Not vectors but its a different way.

 

[RohitShubesh21]

Is this essentially Thale's theorem? ... here's how I proved it without vectors ...

Yes this one I looking for also.

Attempt this:

View attachment 33590

Ok I will Try. Thank You.

iirc there's a way to show this through showing that the vectors that inscribe the semi-circle are perpendicular through the normal equation. Not vectors but its a different way.

How? Can you Show?

 

[ExtremelyBoredUser]

How? Can you Show?

Just consider that the point D would be cos(theta), sin(theta) since its on the circle then just use the gradient formula for each lines and then multiply both gradients... the rest is self-explanatory.

 

[RohitShubesh21]

Just consider that the point D would be cos(theta), sin(theta) since its on the circle then just use the gradient formula for each lines and then multiply both gradients... the rest is self-explanatory.

View attachment 33592

ók I Will Try thanks.

 

[Drongoski]

Just consider that the point D would be cos(theta), sin(theta) since its on the circle then just use the gradient formula for each lines and then multiply both gradients... the rest is self-explanatory.

View attachment 33592

Using ExtremeBoredUser's diagram above, this is a vector method:

CD = AD - AC

BD = AD - AB

AB = -AC

.: CD.BD = (AD - AC).(AD - AB) = (AD - AC).(AD + AC) = AD.AD + AD.AC - AC.AD - AC.AC = AD.AD - AC.AC

= r^2 - r^2 = 0 (r being the radius)


.:, since the dot product CD.BD = 0, CD and BD are perpendicular to each other.


Sorry for not using proper vector notation.

 

[CM_Tutor]

Sorry for not using proper vector notation.

Actually, though it is not used as much, bolding is an accepted vector notation.

 

[CM_Tutor]

The converse of this theorem, that if is a right angle then is the diameter of the circle on which also lies can also be done by vector methods. The question has often been put as:

Let be right-angled at . Use vectors to show that the midpoint of is equidistant from all three vertices of the triangle.

 

[RohitShubesh21]

Using ExtremeBoredUser's diagram above, this is a vector method:

CD = AD - AC

BD = AD - AB

AB = -AC

.: CD.BD = (AD - AC).(AD - AB) = (AD - AC).(AD + AC) = AD.AD + AD.AC - AC.AD - AC.AC = AD.AD - AC.AC

= r^2 - r^2 = 0 (r being the radius)


.:, since the dot product CD.BD = 0, CD and BD are perpendicular to each other.


Sorry for not using proper vector notation.

yes this one sir thank you

 

[Drongoski]

yes this one sir thank you

You're only in Yr 9? Wish I was as smart as you when I was your age.