ABCD is a square of side length 2. A semicircle is drawn on side AB. The line CE is tangent to the semicircle and intersects AD at E.
Find the length of EX.
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Let E X = x .
Since tangents to a circle have the same length, thus E A = E X = x and C X = C B = 2 .
Applying Pythagorean theorem on triangle D E C ,
D E 2 + D C 2 = C E 2 ⇒ ( 2 − x ) 2 + 2 2 = ( 2 + x ) 2
Expanding and simplifying, we obtain 4 = 8 x or that x = 2 1 .