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Altitude in geometry proofs

altitude in geometry proofs

Using the standard notations, in Delta ABC, there are three altitudes: AH_a, BH_b, CH_c.
Extend AH beyond.Therefore, we get three quadrilaterals inscribable in a circle: BH_cHH_a, BH_cH_bC, and CH_bHH_a.How do you prove it?Assume H is the point of intersection of AH_a and BH_b.If you're seeing this message, it means we're having trouble loading external resources on our website.AMS 2008, isbn,. .
Springer, 2009, isbn,.
The triangles are therefore similar.
You may see "the altitude of the triangle is 3 centimeters".
Beginalign CH_c: y - y_c -(x_a - x_b)cdot x (x_a - x_b)cdot x_c / (y_a - y_b) AH_a: y - y_a -(x_b - x_c)cdot x (x_b - x_c)cdot x_a / (y_b - y_c) BH_b: y - y_b -(x_c - x_a)cdot x (x_c - x_a)cdot x_b.Endalign Equations of the 3 side lines are beginalign c AB: y - y_a (y_a - y_b)cdot x x_acdot y_b - x_bcdot y_a / (x_a - x_b) a BC: y - y_b (y_b - y_c)cdot x x_bcdot y_c - x_ccdot y_b / (x_b - x_c).By symmetry usb autorun creator 1.0.21 (or considering triangles AHC and BHK) we get HBcdot HC HAcdot.In other words, the lines a and b intersect on the normal h_c to the line c dropped from the point O and the points C, C_b, and C_a are identical.Again, we have to set up only 1 equation, the other 2 are given by the cyclic permutation of A, B,.Respectively: angle H_cBHangle H_cH_aH, angle H_cBH_bangle H_cCH_b, and angle H_bCHangle H_bH_aH.This means that ACcdot HB0 and AHcdot BC0, such that also ABcdot CH0.Let the coordinates of the 3 vertices be: beginalign A (x_a, y_a B (x_b, y_b C (x_c, y_c).Define H via OH OA.Orthocenter as Incenter In Delta ABC, Delta H_aH_bH_c is known as the orthic triangle.The other two angles are treated similarly.Slope of AB is -a/b hence slope of CC' is b/a and equation of CC' is y (b/a)cdot (x - c) (b/a)x -(bc/a).(A more general statement appears as Theorem 184 in A Treatise On the Circle and the Sphere.For example, beginalign AHcdot BC (AO OH)cdot (BO OC) (-OA OA OB OC)cdot (OC - OB) (OB OC)cdot (OC - OB) OCcdot OC - OBcdot OB OC2 - OB2 0, endalign because O is the circumcenter of Delta ABC.