F.5 Maths!!!

6.
EC = (16 cm) - DE (given)
ED = (16 cm) - AE (given)

DE x EC = AE x EB (intersecting chords)
DE x [(16 cm) - DE] = AE x [(16 cm) - AE]
Hence, DE = AE

DE = AE (proven)
OE = OE (common side)
OD = OA (same radii)
ΔODE ≅ ΔOAE (SSS)

∠OEA = ∠OED (corr. ∠s, ΔODE ≅ ΔOAE)
But ∠OEA + ∠OED = 90° (given)
Hence, ∠OED = 45°

Draw OH⊥DE, and cut DE at H.

In ΔOHE :
∠OED = 45° (proven)
∠OHE = 90° (construction)
∠HOE = 180° - (45° + 90°) = 45° (∠sum of Δ)
Hence, ΔOHE is an isos. Δ with HE = OH (equal ∠sto equal sides)

OE² = OH² +HE² (Pythagorean theorem)
(6√2 cm)² = OH² +OH²
OH = 6 cm

In rt. ∠-ΔOHD :
OD² = OH² +HD² (Pythagorean theorem)
OD² = (6 cm)² +(16/2 cm)²
OD = 10 cm

The radius of the circle = OD = 10 cm

2014-07-07 19:35:53 補充：
There is a typo in line 2. It should be :
EB = (16 cm) - AE (given)