Question

Let 4ABC be inscribed in a circle with center O with AB = 13, BC = 14, AC = 15. Let the foot of the perpendicular from A to BC be D and let AO intersect BC at E. Given the length of DE can be expressed as m/n where m, n are relatively prime positive integers, find m + n.

A. 9/5
B. 25/7
C. 12/7
D. 15/7

Answer 1

To find the length of DE, use properties of right triangles and medians to deduce that DE = 15/7, which simplifies to 15 + 7 = 22.

Step-by-step explanation:

To solve this problem, first, recognize that triangle ABC is inscribed in a circle, making it a right triangle according to the converse of the Thales' theorem (the hypotenuse AC will subtend a right angle at D). By the Pythagorean theorem, the fact that AB = 13, BC = 14, and AC = 15 indicates that triangle ABC is a 5-12-13 multiplied by 3 (right triangle), with the right angle at B.

Now, we can use the fact that AD is the altitude of a right triangle from the right angle to the hypotenuse: AD = (AB * BC) / AC. By substituting the known lengths, we get AD = (13 * 14) / 15.

Since AO is the radius of the circumcircle, and triangle ABC is right, AO is the median to the hypotenuse of the triangle, therefore AO is half of AC. Hence, AO = AC / 2 = 15 / 2.

Since E is the intersection of AO and BC, and D is the foot of the perpendicular from A, DE = AO - AD. Therefore: DE = AO - AD. By calculating the respective lengths, we get DE = (15/2) - ((13 * 14) / 15), which simplifies to DE = 15/7.

The final step is to express DE = 15/7 as the irreducible fraction m/n where m = 15 and n = 7, and finding m + n gives us 15 + 7 = 22.