This time I’m going to prove a geometric dilemma. Just like my 1st proof for the equality between 1 and 2 the result I get is totally absurd. So , here we go……

Theorem: All the triangles are equilateral.

Proof: Let ABC be a general triangle

As indicated in the diagram above, we have constructed some lines such that O is the intersection point of the angle bisector of Â(namely AO) and the perpendicular bisector of BC (namely OZ). X and Y are the feet of the altitudes from O to AB and AC respectively. Z is the midpoint of BC.

From ❶+❷ we get ,

AX+XB=AY+YC

AB=AC

Similarly we can prove that , AC=BC

∴ AB=BC=CA

This means, whatever the triangle ABC is , it is equilateral !!!!.

(For an example if you take a triangle with side lengths 1,2 and 2 then above proof implies that 1=2=2, simply 1=2 😉 )

So , without going any further I leave the challenge of resolving this dilemma to you guys…… 🙂