This is a problem of trigonometry. We will apply the laws of sines and cosines as well as few other trigonometric tricks to solve this problem. First we need to name a few points and segments.
Now we will write down what we are given in mathematical language. We are going to use the trigonometric relationship for area of a tringle, which states area is equal to ab sin(C)/2.
a2 + c2 = b2 + ac sin(x)/2 + ac sin(180-x)/2 + ab sin(180-C1)/2 + bc sin(180-A2)/2
It can be checked with the summation formula for sines that sin(180-t) = sin(t). Therefore,
a2 + c2 = b2 + ac sin(x)/2 + ac sin(x)/2 + ab sin(C1)/2 + bc sin(A2)/2
We can apply the law of sines for the middle triangle which shows b/sin x = a/sin A2 = c/sin C1 Therefore:
a2 + c2 = b2 + ac sin(x)/2 + ac sin(x)/2 + ac sin(x)/2 + ac sin(x)/2
a2 + c2 = b2 + 2ac sin(x).
The law of cosines applied to the middle triangle states a2 + c2 = b2 + 2ac cos(x).
Hence, cos(x) = sin(x)
Thus, x = 45 degrees .
Topic: Trigonometry