∫∠BCD is the common angle of △CDA and △CBD, and CD is the common edge of △CDA and △CBD.

∠∠CDA =∠CBD。

∴△CDA∽△CBD

That is, the similar corresponding sides of two triangles are proportional.

∴

That is, CD2=CA×CB.

(2) It is proved that connecting OD, OD, BO and OA is a radius ⊙ o. ..

Ba is o? The diameter of ∴? ∠BDA=90

∴∠ODA+∠BDO=90

BO = DO，∠CBD=∠CDA。

∴∠CBD=∠BDO=∠CDA

That is ∠ CBD+ODA = 90; Official development assistance+comprehensive development assistance =90.

∴OD⊥CD。 ∵D ∵ O again? There is a little on the table.

∴CD is the tangent of⊙ O.

(3)CDA =∠CBD

∴∠CDA is the tangent angle of the chord AD, and D is the tangent point.

Connecting outer diameter

Then the OD⊥CD.

∠ AOD = 2 ∠ The central angle of CBD is equal to twice the central angle.

∠C+∠AOD = 90°

∠C=90 -2∠CBD=90 -2∠CDA

tan∠C = tan(90-2∠CDA)= 1/tan 2∠CDA

According to the double angle formula? tan2∠CDA=2*2/3？ /( 1-(2/3)? )= 12/5

∴? tan∠C=5/ 12

BE = BCtan∠C = 12 * 5/ 12 = 5