In an acute-angled triangle MNK from point D, the midpoint of side MK, perpendiculars DA

In an acute-angled triangle MNK from point D, the midpoint of side MK, perpendiculars DA and DB are drawn to sides MN and NK, prove that if DA = DB, then triangle MNK is isosceles.

Consider two triangles, MAD and DBK, by condition they are rectangular, since AD ⊥ MN; DB ⊥ NK, angle <MAD = 90 °; <DBK = 90 °, and the hypotenuses MD and DK are opposite the right angle.

Let us calculate the legs of the triangles MA and BK, taking into account that MD = DK by condition. Then:

AM ^ 2 = MD ^ 2 – AD ^ 2 in AMD triangle; BK ^ 2 = DK ^ 2 – DB ^ 2 in triangle BDK. Even without extracting the root, we conclude that AM ^ 2 = BK ^ 2, since they contain MD = DK in the formulas; and AD = BD by condition. We got that triangles AMD and DBK are equal, angles <AMD = <BKD are equal, triangle MNK is isosceles.