F.4 maths quesitons help!!!

1a. Since BD and BE are tangents,
thus we have ∠OBA=∠OBC=30 degree (Tangent Properties)

Similarly, since CE, CF are tangents,
we have ∠OCB=∠OCA=24 degree (Tangent Properties)

Consider ΔABC,
∠DAF + 2*30 degree + 2*24 degree = 180 degree (∠ sum of Δ)
thus ∠DAF=72 degree.

b. Since AD and AF are tengents, AD=AF (Tangent Properties)
therefore ∠ADF=∠AFD (base ∠s, isos. Δ)

Now consider ΔADF,
72 degree+∠ADF+∠AFD = 180 degree (∠ sum of Δ)
thus ∠ADF=54 degree.

2a. [POB is a tangent line? Is there a typo? ]

Since PC is a tangent, thus OC⊥PC (tangent ⊥ radius)
that means ∠OCP=90 degree.

Now consider ΔOCP,
30 degree + 90 degree + ∠POC = 180 degree (∠ sum of Δ)
∠POC=60 degree.

b. Consider ΔOBC, since OB=OC (radii)
thus ∠OBC=∠OCB (base ∠s, isos. Δ)
Also, ∠OBC+∠OCB=60 degree (ext. ∠ of Δ)
therefore we have ∠OBC=∠OCB=30 degree.

On the other hand, since PA, PC are tangents,
thus ∠OPA=30 degree (Tangent properties)

Now we have already shown that ∠OBC=∠OPA
Hence, CB//PAE (alt. ∠s eq.).

c. Since PA, PC are tangents, PA=PC (tangent properties)
Thus ∠CAP=∠PCA (base ∠s, isos. Δ)

Consider ΔPAC,
60 degree + ∠CAP+∠PCA=180 degree (∠ sum of Δ)
Therefore ∠CAP=60 degree.

2009-01-02 16:14:12 補充：
d. By (b), CB//PAE. By symmetry, AB//PCF.
Thus PABC is a //gram (definition).

Since ∠OPC=30 degree, thus ∠OPA=30 degree (tangent properties)
therefore ∠APC=60 degree.
Hence, ∠ABC=60 degree (opp. ∠s, //gram)

2009-01-02 16:14:20 補充：
On the other hand, since CB//PAE,
∠PAB+60 degree = 180 degree (int. ∠s, CB//PAE)
thus ∠PAB=120 degree and hence ∠CAB=60 degree.
Similarly, ∠ACB=60 degree.

Hence, ΔABC is equilateral triangle.

1

Draw OD,OF, OE

Then



















2008-12-31 15:57:28 補充：
2

(a)