Problem
Given the triangle (sketched below) find $\angle B, \angle D$, the length of the diagonal $BD$ and the area of the quadrilateral.
Given values:
$\angle CAD = 40^o$
$\angle BAD = 90^o$
$CD = 5.0$
$AD = CD$
$AC = AB$
My progress
I found $\angle ACD = 40^o$ by the law of sines.
I found $\angle D = 100^0$ by the angle sum of a triangle, $\angle D = 180 – 2\cdot40$.
The rest remains a mystery to me. It seems like $\angle B$ should be solvable, but I can't see it.
Best Answer
You have that $AB=AC$ so $\angle ABC=\angle ACB$ and once $\angle BAC =50°$ then
$$2\angle ABC+50°=180° \to \angle ABC=65°$$
In order to find the area you just need to calculate $y$.
For that you can use cossine law at $\Delta ACD$:
$$AC^2=y^2=5^2+5^2-2\cdot5\cdot5\cdot \cos 100°\to y^2=25(1-2\cos 100°)$$
and the area will be:
$$(ABCD)=(ABC)+(ACD)=\frac{5\cdot 5\cdot \sin 100°}{2}+\frac{y\cdot y\cdot \sin 50°}{2}$$