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]

misolotwhattheheck Jul 14, 2024

#1**0 **

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.

magenta Jul 14, 2024