A circle is inscribed in an isosceles triangle ABC with a base AC and perimeter 13, k is the point

A circle is inscribed in an isosceles triangle ABC with a base AC and perimeter 13, k is the point of contact of this circle with the side BC. Find the base of the AC if BK = 6.

Let’s draw the height BH, it will pass through the point O (the center of the inscribed circle), since the triangle is isosceles.

The perimeter of the triangle ABC is 13 cm, which means that the sum of the two segments BC + CH is 13/2 = 6.5 cm (since AB = BC and the height BH is the median, CH is half of the AC).

Consider triangles KOС and COH: both triangles are rectangular, OK = OH (circle radii), OS is the common side. So the triangles are equal.

Therefore, KC = HC. We denote this length by X.

Since BC + CH = 6.5, and BC + CH = 6 + x + x, the equation is obtained:

6 + 2x = 6.5.

2x = 6.5 – 6;

2x = 0.5;

x = 0.5 / 2 = 0.25.

The base of the AC is twice the length of the CH segment, it turns out that AC = 0.25 * 2 = 0.5 cm.

Answer: The base of the AC triangle is 0.5 cm.