數學三角相關定理

12 ∠BAC = 180 - ∠ABO - ∠OBC - ∠ACO - ∠OCB= 180 - 2∠OBC - 2∠OCB∠BOC= 180 - ∠OBC - ∠OCB= 90 + (1/2)∠BAC13 ∠ACB= 180 - ∠BAF - ∠FAC - (180 - ∠EBD - ∠DBF)= ∠EBD + ∠DBF - ∠BAF - ∠FAC∠ADB= 180 - ∠BAF - (180 - ∠EBD - ∠DBF) - ∠DBF= ∠EBD - ∠BAF=(1/2)∠ACB14 AD = AD∠ACD = ∠AED∠EAD = ∠DAC三角形ACD全等三角形AED

DC = DE

(b) 三角形BDE周界

= BD + DE + BE

= BD + DC + BE

= BC + AC

= 2√32

= 8√2

Suppose that I have solve Q14a.
____________
BC^2+BC^2=8^2(Pyth. thm.)
BC=4sqrt2
___________
BD+DE=4sqrt2
__________________
BE=8-4sqrt2
________________
8-4sqrt2+4sqrt2
=8cm

第12題點解
∠BOC

= 180 - ∠OBC - ∠OCB

= 90 + (1/2)∠BAC


第13題點解

= ∠EBD - ∠BAF

=(1/2)∠ACB


第14題答案應要是8 點計 ?