問兩題圓形

26.

(a)
In △PBC and △PDA,
∠BPC = ∠DPA (common angle)
∠ABC = ∠ADC (∠s in the same segment)
i.e. ∠PBC = ∠PDA
∠PCB = ∠PAD (3rd ∠ of △)
∴ △PBC ~ △PDA (equiangular)

(b)
∠DPT = ∠PBC (∠ in alt. segment)
∠PBC = PDA (proved)
∴ ∠DPT = ∠PDA
∴ AD // PT (alt. ∠s eq.)

---

27.

(a)
唔識 ...

(b)
∠TCA = ∠CBT (∠ in alt. segment)
∠TAC = ∠CBT (∠s in the same segment)
∴ ∠TCA = ∠TAC
∴ TA = TC (sides opp. eq. ∠s)

(c)
∠TCA = ∠CBT (∠ in alt. segment)
∴ ∠CBT = 30°
∠TCA = ∠ABT (∠s in the same segment)
∴ ∠ABT = 30°
∠ABC = ∠ABT + ∠CBT
∠ABC = 30° + 30°
∠ABC = 60°

JY4066 :27.

(a)
唔識 ...

THANKS
27a 啊sir教過...系 tangent⊥radius 之後用 90°+90° = 180° (opp ∠ supp.)

GM7811 - 834 @ N8x :

JY4066 :27.

(a)
唔識 ...

THANKS
27a 啊sir教過...系 tangent⊥radius 之後用 90°+90° = 180° (opp ∠ supp.)

But ,
there are no radii ...