Proof1 (Dinesh Jagai)

Problem 1)
https://www.geogebra.org/m/MAdsQxwT
In the figure, we have an arbitrary triangle ABC with altitudes AHa , CHc , BHb.

Consider the following image above in GeoGebra. In order to prove Pythagoras theorem, I shall prove a stronger statement:

Namely, AB * BHc + AC * CHC = BC2

Define angles a,b,c to represent angles A,B,C respectively. Now,in triangle BCHb BHc = BCcosb ,since angle Hc is a right angle. Recall lines CHc ,BHb and AHa are altitudes.Similiarly, CHb = BCcosc , BHa = ABcosb and HaC = ACcosc.

Now we have AB*BHc + AC*CHb = AB*BCcosb + AC*BCcosc
= BC*(AB.cosb + AC.cosc)
= BC*(BHa + CHA)
= BC * BC
= BC2
We have just proven the general statement.Now if Angle A is a right angle, the altitude BHb will be the line AB and the altitude CHC will be the line AC. Thus, AB = BHc and AC = CHb
This leads up to AB2 + AC2 = BC2, which is the familiar Pythagoras theorem!

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