# A rhombus with a side of 6 cm is inscribed in a right-angled triangle with an angle of 60 * so that this angle is common

**A rhombus with a side of 6 cm is inscribed in a right-angled triangle with an angle of 60 * so that this angle is common to them and all the vertices of the rhombus lie on the sides of the triangle. In what respect does the apex of the rhombus divide the larger leg of the triangle?**

Consider a right-angled triangle DВK in which the angle ВDK = BAC = 60 as the corresponding angles at the intersection of parallel lines DC and AC secant AB. Then the angle DK = 180 – 90 – 60 = 30. The leg of the DK lies opposite the angle 30, therefore, its length is equal to half the length of the hypotenuse of the ВD, then ВD = 2 * DK = 2 * 6 = 12 cm.By the Pythagorean theorem, ВK ^ 2 = ВD ^ 2 – DK ^ 2 = 144 – 36 = 108.

ВK = 6 * √3 cm.

In the ADKM rhombus, we draw the height of the BP, then, in the right-angled triangle ADН, the angle ADН= 180 – 90 – 60 = 30, then AH = AD / 2 = 6/2 = 3 cm.

By the Pythagorean theorem, DН ^ 2 = AD ^ 2 – AH ^ 2 = 36 – 9 = 27. DН = 3 * √3 cm.

Then KС = 3 * √3 cm.

Let’s define the ratio ВK / KС = 6 * √3 / 3 * √3 = 2.

Answer: The top of the rhombus divides the leg in a ratio of 2/1.