Hi everyone,
I need some help.
How many Thales Theorem Converse (inscribe Circle) are there?
Can anyone show me different proofs of Thales Theorem Converse (have to be inscribe Circle)
Thanks a lot.
I am very confused with this one
Hi everyone,
I need some help.
How many Thales Theorem Converse (inscribe Circle) are there?
Can anyone show me different proofs of Thales Theorem Converse (have to be inscribe Circle)
Thanks a lot.
I am very confused with this one
OK what does Thales Thrm say. If A, B, C are points on a circle, with AB being the diameter then the triangle ABC is a right triangle. So the converse is If we have a right triangle ABC, then AB is the diameter of its circumcircle.
Here is a neat proof using linear algebra.
Refer to the diagram
Now Lets say we have a right triangle ABC, Let AB represent its longest side (Hypotenuse). Now Make a circle with diameter AB with the center of the line AB being the "origin".
Now we need to show that point C also lies on this new circle with diameter AB.
Note that the vector (A-C) is perpendicular to the vector (C-B) . So $\displaystyle (A-C) \dot (C-B) = 0 $ . Remeber dot product of perpendicular vectors is 0. So we can distribute this through. we get $\displaystyle AC - AB - CC + CB = 0 $ . Now since AB is the diameter and the origin is the center of AB. The Vector A = -B .So substituting $\displaystyle AC + AA - CC - CA = 0 $ which equals $\displaystyle AA - CC = 0 $ since for any vector dot product by itself is equal to its magnitude square $\displaystyle a \dot a = ||a||^2 $, we get $\displaystyle ||A||^2 = ||C||^2 $ or $\displaystyle ||A|| = ||C|| $ , since ||A|| = radius, this means ||C|| = radius, so C lies on the circle and thus the circle we made with AB as the diameter is indeed the circumcircle for the right triangle ABC. QED.
Here is a geometric proof, the linear alg proof is a bit confusing if ur not too familiar with vectors.
Assume ABC is a right triangle, Now Draw the circumcircle containing the points ABC. I will try to show that the longest side of our triangle ABC corresponds to the diameter of this circumcircle. (Attached Pic 1)
Now draw a line parallel to the segment CB passing through point A. (Attached Pic 2)
Now draw a line parallel to the segment AC passing through point B. (Attached Pic 3)
Now the parallelogram formed ABCD. (Attached Pic 4 For the rest )
Since the angle ACB is 90 Deg ADB is also 90 Deg.
It is also easy to see CAD is 90 deg and therefore CBD is also 90 deg.
Since in our paralleogram all the angles are 90. It is a rectangle.
Since the diagonals in a rectangle A) intersect at their median point. B)are equal.
We see that the intersection, label it O (for origin), the segments OC, OB, OA are all equal in length. Since A, B, C lie on the circle, this means that O is the origin of the circum circle and since AB was one of the diagonals of our rectangle, AB is the diameter in the circumcircle.
QED
yea! great, i fully understand this theorem by the geometric proof.
I m working on a project about thales thm converse and need to show different kinds of proofs(as much as possible)
but i google online, most of them r geometric or linear algebra proofs.. do u think there are other as well?
Thanks for the proof
How about this. Prove that the midpoint of the hypotenuse of a right triangle is equidistant from all the verticies. Since all three points lie on the circumcircle containing the triangle ABC, it implies that the midpoint of the hypotenuse of the triangle is the origin of the circumcircle, and since the hypotenuse passes through it (obviously), the hypotenuse is the diameter.
Here, Assume that we have a right triangle ABC and its circumcircle. Now i will prove that the midpoint of the hypotenuse AB is equidistant from all the vertices of the triangle .
See the picture, its easy to see that BGO and ABC are similar triangles. which means if ABC has a side ratio A:B:C then BGO has $\displaystyle \lambda A_1 : \lambda B_1 : \lambda C_1 $. Since O is the midpoint of the hypotenuse, OB = $\displaystyle \frac{c}{2} $ or 2*OB = c, which means 2 is our $\displaystyle \lambda $. Which means that $\displaystyle \frac{B}{2} = B_1 $ and $\displaystyle \frac{A}{2} = A_1 $, which means $\displaystyle (A_1)^2 + (B_1)^2 = (C_1)^2 $ = $\displaystyle \frac{1}{4} (A^2 + B^2) = (C_1)^2 $ so $\displaystyle \frac{1}{4}{c^2} = (c_1)^2 $ and so $\displaystyle c_1 = \frac{c}{2} $ .Which means all verticies are $\displaystyle \frac{c}{2} $ distance away from the midpoint of the hypotenuse of the right triangle, since ABC lies on the circum circle, this imples the midpoint of the hypotenuse is the origin of the circumcircle and thus the hypotenuse is the diameter of the circumcircle
But that does show that a right triangles hypotenuse is the diameter of its circumcircle.
It shows that the midpoint of the right triangles hypotenuse is the same distance from all its points. Since the points A,B,C lie on its circumcircle, What is the only point in a circle which was the same distance to points on the circle? The origin. This implies that the midpoint of the hypotenuse of the right triangle is the origin of the circumcircle containing A,B,C. Since the hypotenuse goes through the midpoint (origin of the circle), it must be the diameter of the circumcircle.