In the following figure prove that: AK||BCAK||BC

In arbitrary triangle ABCABC, let GG be centroid and AHAH be a height.We extend HGHG beyond GG to intersect circumcircle at KK.Prove that AK||BCAK||BC

I connected midpoint of BCBC to KK and AA to show MA=MKMA=MK but failed…

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