Triangle ABC is circumscribed about circle O and D,E, and F are points of tangency. If AD= 5, EB= 5 and CF= 10, find the lengths of the sides of the triangle and show that the triangle is isosceles

To find the lengths of the sides of triangle ABC, we can observe that the lengths of the tangents drawn from the vertices of a triangle to a circle are equal.

We are given that AD = 5, EB = 5, and CF = 10. These lengths represent the tangents from points A, B, and C respectively.

Now let's analyze the triangle:

1. AD = 5: This is the tangent from point A. Since AB is a tangent to the circle, we can conclude that AB = AD = 5.

2. CF = 10: This is the tangent from point C. Since CB is a tangent to the circle, we can conclude that CB = CF = 10.

3. EB = 5: This is the tangent from point B. Since BC is a tangent to the circle, we can conclude that BC = BE = 5.

From the above analysis, we can see that AB = AD = 5, BC = BE = 5, and CB = CF = 10.

Therefore, triangle ABC is isosceles, as it has two sides (AB and BC) of equal length.

To find the lengths of the sides of triangle ABC, we need to understand some properties of a triangle circumscribed about a circle.

1. Tangent segments drawn from the same external point to a circle are equal in length. This means that AD = AE, BE = BF, and CF = CD.

2. In a triangle, if two sides are equal, then the opposite angles subtended by these sides are also equal.

Now let's use these properties to find the lengths of the sides of triangle ABC:

Since AD = AE and CF = CD, we can replace AD with AE and CF with CD to simplify calculations.

Let's denote AD (or AE) as x, BE as y, and CF (or CD) as z.

From the given information, we have:
x = 5 (AD = 5)
y = 5 (EB = 5)
z = 10 (CF = 10)

Using property 1, we know that x = y + z. So, we can rewrite the equation as:
5 = 5 + z

Subtracting 5 from both sides, we get:
z = 0

Now, using the equation x = y + z, we substitute the values of x and z to find y:
5 = y + 0
y = 5

So, we have:
x = 5
y = 5
z = 0

Now, we can use the given information AD = 5, EB = 5, and CF = 10 to find the lengths of the sides of triangle ABC:

AB = AD + BE = 5 + 5 = 10
BC = BE + CF = 5 + 10 = 15
AC = AD + CF = 5 + 10 = 15

Therefore, the lengths of the sides of triangle ABC are:
AB = 10
BC = 15
AC = 15

To show that the triangle is isosceles, we need to prove that at least two sides of the triangle are equal.

From the calculations above, we see that AC = BC = 15, so we can conclude that triangle ABC is isosceles.

shown that B'=90 degrees