In the right-angled triangle MNK with the hypotenuse NK, the bisector KD and the perpendicular DE

In the right-angled triangle MNK with the hypotenuse NK, the bisector KD and the perpendicular DE to the hypotenuse are drawn. Prove that if NE = EK then MN = 3MD.

The two corners of the MDB triangle are equal to the two corners of the KDE triangle:

<DMK = <DEK = 90 °.

<MKD = <DKE (formed by the bisector of angle K).

These triangles also have one side in common, KD.

Conclusion: ΔMDB = ΔKDE.

MK = KD; NK = 2KD (by condition).

NK = 2MK.

NK / MK = 2;

The bisector KD divides the MN side in the relation in which the adjacent NK and MK sides are located:

DN / MD = NK / MK = 2;

DN = 2MD

MN = MD + DN = MD + 2MD = 3MD.

Answer: MN = 3MD.