/sci/ - Science & Math

>>10126492
no

(wlog assume an unit circle)
imagine a circle, the point on the circumference is (x, y)

split the triangle with the line that passes at (x, 0) and (x, y)

this forms 2 triangles:
(-1, 0), (x, 0), (x, y) [ABC]
(1, 0), (x, 0), (x, y) [DEF]

so what you're really asking is if angle(CA, CB) + angle(FE, FD) = 90

the rest should be trivial

>>10126495
yes*

>>10126492
Because the diagonals of rhombus are at right angles.

>>10126492

https://en.m.wikipedia.org/wiki/Thales's_theorem

easy.

Let B and C be the vertices on the diameter's ends, and A the last vertex.

Let sA, sB and sC be the spreads associated with the vertices A B and C.

let Q be the quadrance of the side AB, P the quadrance of the side AC, and D the quadrance of the side BC.

The quadrance of OA, OB and OC is D/4.

In the triangle OAB, using the cross law gives us sB = 1- Q/D.

In the triangle OAC, we get sC = 1- P/D

Now the spread law is sA/D=sB/P=sC/Q

In particular, sB*Q=sC*P.

Using what we found and pluging in, we have the following equality:

Q - Q^2/D = P - P^2/D

ie QD - Q^2 = PD - P^2

so D(Q-P) = (Q-P) *(Q+P).

if Q is not equal to P, we have D=Q+P, which according to Pythagora's theorem means ABC is a right triangle in A.
if Q=P then sB=sC, and OAB and OAC are right trianglss in O.

Apply the cross law in OAB for example (D/4 +Q-D/4)=(1-sB) *4*Q*D/4

Therefore Q=D/2. Same for P. Thus D=P+Q, leading to the same conclusion: ABC is a right triangle in A.

>>10126492
Yes.
Draw a circle and lay a diameter line on it from points A to B. Mark the center O of the circle. Chance upon a point C of the triangle, and connect the triangle ABC. Also connect OC.
Since a circle is a set of points at the same distance of the center, OA, OB and OC have the same measure. Since the two triangles are isosceles, OAC and OCA have the same angle, and similarly do OBC and OCB. But BOC and AOC sum two right angles, and the whole of a triangle also sums two right angles, so the remaining angles of the two triangles, OAC, OCA, OBC and OCB, sum two right angles. Since OAC is equal to OCA, and OBC is equal to OCB, OCA and OCB sum a right angle.

Man I love Euclidposting.
>>10126495
>analytic geometry
No, friend.
>>10126533
What the fuck even is this.

>>10126654
>chance upon a point C of the triangle
Of the circle, sorry.

>>10126654
>>10126533
>What the fuck even is this.
Rational trigonometry.

>>10126497
you had one job...