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=252−25=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.
