# In an isosceles triangle with a base of 12cm and a height of 8cm, a square is inscribed so that two

**In an isosceles triangle with a base of 12cm and a height of 8cm, a square is inscribed so that two of its vertices lie on the base of the triangle, and the remaining two are on its lateral sides. Find the side of the square. The answer is 6, 12.**

Let a triangle ABС be given, BН = 8 cm – the height of the triangle, K – the point of intersection of the square with the side AB, L – the point of intersection of the BН with the side of the square.

Let’s denote X – the side of the square.

Obviously, the triangles ABN and KBL are similar, since the ABN has equal angles with ABН.

Let’s write down the property of proportionality of such triangles:

BН / BL = AН / CL = k.

AH = AC / 2 = 6, CL = X / 2, BL = 8 – X.

Then,

8 / (8 – X) = 6 / (X / 2).

8 * X = 12 * (8 – X).

8 * X = 96 – 12 * X.

20 * X = 96.

X = 4.8 (cm).

The side of the square is 4.8 cm.