In triangle ABC, if the medians AX and BY are mutually perpendicular such that BC2 + AC2=k2(AB2) , where k is a real number , then the value of k can be  what? 

Dear student

Given: AX and BY are medians.Now ,In AOB,By Pythagoras Theorem, we haveAB2=OB2+OA2   ...(1)In BOX,By Pythagoras Theorem, we haveBX2=OB2+OX2   BC24=OB2+OX2     As AX is a median on BCOB2=BC24-OX2     ...(2)In AOY,By Pythagoras Theorem, we haveAY2=OA+OY2   AC24=OA2+OY2     As BY is a median on ACOA2=AC24-OY2     ...(3)In XOY,By Pythagoras Theorem, we haveXY2=OX2+OY2   ...(4)Consider eq(1), we haveAB2=OB2+OA2 AB2=AC24-OY2  +BC24-OX2 AB2=122AC2+BC2-OX2+OY2    [using 2 and 3]122AB2+XY2=AC2+BC2     [using 4]Please check your question.There is some mistake.
Regards

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