# In a right-angled triangle ABC (angle C = 90 degrees), the height CD is drawn so that the length of the segment BD

**In a right-angled triangle ABC (angle C = 90 degrees), the height CD is drawn so that the length of the segment BD is 4 cm longer than the length of the segment CD, AD = 9cm. Find the sides of the triangle ABC.**

According to the property of the height of a right-angled triangle drawn from the vertex of the right angle, the height is the average proportional to the projections of the legs on the hypotenuse.

СD = √AD * ВD.

Let the length of the height СD = X cm, then, by condition, the length of the segment is equal to: ВD = (X + 4).

Then: X = √ (9 * (X + 4)) = √ (9 * X + 36).

Let’s square both sides of the equality.

X ^ 2 = 9 * X + 36.

X ^ 2 – 9 * X – 36 = 0.

Let’s solve the quadratic equation.

D = b ^ 2 – 4 * a * c = (-9) ^ 2 – 4 * 1 * (-36) = 81 + 144 = 225.

X1 = (9 – √225) / (2/1) = (9 – 15) / 2 = -6 / 2 = -3. (Doesn’t fit because <0).

X2 = (9 + √225) / (2/1) = (9 + 15) / 2 = 24/2 = 12.

CD = 12 cm, then DB = 12 + 4 = 16 cm.

AB = AD + DB = 9 + 16 = 25 cm.

From the right-angled triangle ACD we define the hypotenuse AC.

AC ^ 2 = SD ^ 2 + AD ^ 2 = 144 + 81 = 225.

AC = 15 cm.

From the right-angled triangle ABC, we define the leg BC according to the Pythagorean theorem.

BC ^ 2 = AB ^ 2 – AC ^ 2 = 625 – 225 = 400.

BC = 20 cm.

Answer: AB = 25 cm, BC = 20 cm, AC = 15 cm.