About circle 5

19b) Join AD and DC, then ∠TDA = ∠DCA (∠ in alt. segment)

Also since CFDE is a cyclic quad., ∠DFE = ∠DCE (∠s in the same segment)

Hence ∠DFE = ∠TDA

So AD//EF

22b) ∠PQA = ∠QBA and ∠PRA = ∠RBA (∠s in alt. segment)

Also ∠PQA = ∠RBA and ∠PRA = ∠QBA (∠s in the same segment)

Thus ∠PQA = ∠PRA and hence △PQR is an isos. △

So PQ = PR

2010-07-27 23:58:24 補充：
Let me think for a while more for Q23. Thanks.

2010-07-28 22:36:23 補充：
23a) By tangent prop., we have DC = DB. Thus ∠DCB = ∠DBC and we let them be θ.

So ∠BAC = θ (∠ in alt. segment)

Now, we let ∠DCE = ∠CFE = Φ (∠ in alt. segment)

Then ∠CDE = 2θ (ext. ∠ of triangle)

So, ∠CED = 180 - 2θ - Φ

2010-07-28 22:37:42 補充：
Also, ∠CED = ∠CFE + ∠ECF (ext. ∠ of triangle)

180 - 2θ - Φ = Φ + ∠ECF

∠ECF = 180 - 2θ - 2Φ

∠ACB = 180 - ∠ECF - ∠ECB

= 180 - (180 - 2θ - 2Φ) - (θ + Φ)

= θ + Φ

= ∠ECB

Hence, △ABC ~ △BEC (AAA)

2010-07-28 22:38:29 補充：
b) With △ABC ~ △BEC:

BC/EC = AC/BC

BC^2 = 16

BC = 4 cm