Draw radii OB and OC, then ∠ BOC = 2 ∠ BAC = 90? ; 2OB？ =(6+4)? (Pythagorean Theorem), ∴ ob = 5 √ 2.

(2)∫AE de = 6×4 (chord-crossing theorem) ... a;

∫∠CAD = 45 again? ÷(6+4)×4= 18? (Inference of the theorem of intersecting chords; The circumferential angles on the same arc CD are equal),

Similarly, ABC = (180? ﹣45? )÷(6+4)×4=54? = 3 ∠ CAD，AE = 3de...b；

By solving equations A and B, DE = 2 √ 2, AE = 6 √ 2 and AD = 8 √ 2 are obtained.

So quadrilateral ABCD area = △ Abd area +△ ACD area =? 8√2(BE+DE)=4√2× 10

=40√2。