Problem: In quadrilateral ABCD with diagonals AC and BD, intersecting at O,BO=4, OD=6,AO=8,OC=3, and AB=6. The length of AD is:
Answer Choices:
A. 9
B. 10
C. 63​
D. 82​
E. 166​
Solution:
82=h2+(4+t)2,62=h2+t2,t=23​
∴h=2315​​∴x2=h2+DE2=4135​+4529​=166∴x=166​.
or
△BOC∼△AOD with side ratio 1:2∴BC=21​x
△AOB∼△DOC∴CD=421​. Since ABCD is inscriptible (why?) we may use Ptolemy's Theorem.
∴x⋅2x​+6⋅421​=(6+4)(8+3)∴x=166​.
