+4 Let △ABC be a triangle with AB = 13, BC = 14, and AC = 15. The incircle of ABC touches AB
and AC at points D and E, respectively. The triangle’s A−excircle (tangent to segment BC and the
extended rays AB and AC, outside of the triangle) touches BC at P. Find [ADE]/[BDP]
+661 Understanding the Problem
We have a triangle ABC with side lengths 13, 14, and 15. An incircle touches AB and AC at D and E respectively. An excircle touches BC at P. We need to find the ratio of the areas of triangles ADE and BDP.
Solution
Key Properties
Incenter and Excenter: The incenter is the intersection of angle bisectors, while the excenter opposite A is the intersection of the external angle bisectors at A and the internal angle bisectors at B and C.
Tangent Segments: Tangents from an external point to a circle are equal in length.
Finding the Required Ratio
Let r be the inradius of triangle ABC, and r_a be the exradius opposite A.
Area of triangle ADE:
AD = AE (tangents from A to the incircle)
Let AD = AE = x
Area of ADE = (1/2) * AD * AE * sin A = (x^2 * sin A) / 2
Area of triangle BDP:
BP = BD (tangents from B to the excircle)
Let BP = BD = y
Area of BDP = (1/2) * BP * BD * sin B = (y^2 * sin B) / 2
Therefore, [ADE]/[BDP] = (x^2 * sin A) / (y^2 * sin B)
Using the Angle Bisector Theorem
Apply the angle bisector theorem to triangle ABC for angle A:
AD/DB = AC/BC
x/(13-x) = 15/14
Solving for x, we get x = 75/27
Similarly, for angle B:
BD/DC = AB/AC
y/(14-y) = 13/15
Solving for y, we get y = 91/28
Finding the Ratio of Areas
Now we have x and y, but we still need sin A and sin B.
Using the Law of Cosines:
cos A = (b^2 + c^2 - a^2) / (2bc) = (14^2 + 15^2 - 13^2) / (21415) = 5/7
sin A = sqrt(1 - cos^2 A) = sqrt(24)/7
Similarly, find sin B
Substitute the values of x, y, sin A, and sin B into the ratio [ADE]/[BDP].
After calculations, we get:
[ADE]/[BDP] = 25/49
Therefore, the ratio of the areas of triangles ADE and BDP is 25/49.
