Nice solution. For those unfamiliar with Power of a Point theorem, we can use Pythagorean theorem by constructing right triangles. Drop perpendicular from A to BC at point P. Since △ABC is isosceles (with AB = AC), altitude AP bisects BC. Let AP = h, BP = CP = a, CD = b. Then BD = 2a + b Using Pythagorean theorem in △APC, we get: h² + a² = 2² (1) Using Pythagorean theorem in △APD, we get: h² + (a+b)² = 4² (2) Subtracting (1) from (2) we get: (a+b)² − a² = 16 − 4 a² + 2ab + b² - a² = 12 2ab + b² = 12 (2a+b) * b = 12 BD * CD = 12
Nice solution.
For those unfamiliar with Power of a Point theorem, we can use Pythagorean theorem by constructing right triangles.
Drop perpendicular from A to BC at point P. Since △ABC is isosceles (with AB = AC), altitude AP bisects BC.
Let AP = h, BP = CP = a, CD = b. Then BD = 2a + b
Using Pythagorean theorem in △APC, we get:
h² + a² = 2² (1)
Using Pythagorean theorem in △APD, we get:
h² + (a+b)² = 4² (2)
Subtracting (1) from (2) we get:
(a+b)² − a² = 16 − 4
a² + 2ab + b² - a² = 12
2ab + b² = 12
(2a+b) * b = 12
BD * CD = 12
You could also get the product by using Stewart's Formula to find the Cevian of a Triangle.