Problem:
Let AB be a diameter in a circle of radius 52​. Let CD be a chord in the circle that intersects AB at a point E such that BE=25​ and ∠AEC=45∘. What is CE2+DE2 ?
Answer Choices:
A. 96
B. 98
C. 445​
D. 702​
E. 100 Solution:
Solution 1 (Pythagorean Theorem) Let O be the center of the circle, and X be the midpoint of CD. Let CX=a and EX=b. This implies that DE=a−b. Since CE=CX+EX=a+b, we now want to find (a+b)2+(a−b)2=2(a2+b2). Since ∠CXO is a right angle, by Pythagorean theorem a2+b2=CX2+OX2=(52​)2=50. Thus, our answer is 2×50=(E)100​.
OR
Let O be the center of the circle, and X be the midpoint of CD. Draw triangle OCD, and median OX. Because OC=OD,OCD is isosceles, so OX is also an altitude of OCD. OE=52​−25​, and because angle OEC is 45 degrees and triangle OXE is right, OX=EX=2​52​−25​​=5−10​. Because triangle OXC is right, CX=(52​)2−(5−10​)2​=15+1010​​. Thus, CD=215+1010​​.
We are looking for CE2+DE2 which is also (CE+DE)2−2⋅CE⋅DE.
Because CE+DE=CD=215+1010​​,(CE+DE)2=CD2=4(15+1010​)=60+4010​.
By Power of a Point, CE⋅DE=AE⋅BE=25​⋅(102​−25​)=2010​−20, so 2⋅CE⋅DE=4010​−40.
