The base (BC) of an isosceles triangle (ABC) is also the side of a square (BCEF) in which the triangle is inscribed. A perpendicular line is drawn in the triangle from B to AC at D, forming triangle ABD. What is the ratio of triangle ABD to the square BCEF?

Answer 1

a graphical solution seems the most straight forward

B(0,0)
C(10,0)
E(10,10)
A(5,10)
F(0,10)

find the equation for AC, and then the equation for BD

D is the intersection of the lines

calculate the lengths of BD and AD to find the area of ABD

Answer 2

Assuming you mean the ratio of areas,

If BC = 2x, then AD = 2x

ABD has area 1/2 * 2x * 2x = 2x^2
ABEF has area (2x)*(2x) = 4x^2

Answer 3

AC ... y = -2 x + 20

BD ... y = 1/2 x

D (8,4)

BD = √(64 + 16)

AD = √(9 + 36)

area ABD = ½ √(80 * 45) = 30

area ABEF = 10² = 100