form 3 math (geometry)

Classwork 10.1
1)Let ∠BGD=z
y=z (alt. ∠s, AB//CD)
∵x=y (given)
i.e. x=z
Hence EB//FD

2)∵∠DEF+∠CDE=150∘+30∘=180∘
∴DC//EF (int. ∠s supp.)
∠BCD=∠CDE=30∘(alt. ∠s, BC//DE)
∵∠ABC+∠BCD=150∘+30∘=180∘
∴AB//DC (int. ∠s supp.)
i.e. AB//EF

Classwork 10.2
1)Construct CG//ED.
∠DCG=∠CDE=70∘(alt. ∠s, CG//ED)
∠BCD=150∘(given)
∠BCG+∠DCG=150∘
∠BCG+70∘=150∘
∠BCG=80∘
∠ABC=80∘(given)
∴∠ABC=∠BCG
i.e. CG//AB


2)Construct CH//AB.
∵CH//AB (by construction)
and AB//DE (given)
∴AB//CH//DE
∠FCH=∠FBA=a (corr. ∠s, AB//CH)
∠GCH=∠FDE=b (corr. ∠s, CH//DE)
∠BCD=∠FCH+∠GCH=a+b

Classwork 10.3
1)∵BC=CD (given)
∴∠CDB=∠CBD (base∠s, isos. △)
∠CDB=∠ABD (alt. ∠s, AB//DC)
∴∠ABD=∠CBD
i.e. BD is the angle bisector of ∠ABC.

2)∵BD=DE (given)
∵∠DBE=∠DEB (∠s, isos. △)
∠DBE=∠EBC (BE is angle bisector.)
∴∠EBC=∠DEB
i.e. DE//BC (alt. ∠s equal)

Classwork 10.4
1)∠ABC=∠DBF=b (vert. opp. ∠s)
∠ACB=∠ECG=c (vert. opp. ∠s)
∠BAC+∠ABC+∠ACB=180∘(∠ sum of △)
a+b+c=180∘

2)∠CED=∠ABC=a (alt ∠s, AB//ED)
∠CED+∠CDE+∠DCE=180∘(∠ sum of △)
a+b+c=180∘

3)∠BAD+∠ADB=∠DBF (ext. ∠ of △)
2x+p=2y
p=2y-2x
∠BAC+∠ACB=∠CBF (ext. ∠ of △)
x+q=y
q=y-x
∵p+q=120∘(given)
2y-2x+y-x=120∘
3y-3x=120∘
y-x=40∘
y=x+40∘

2009-03-25 22:05:25 補充：
Classwork 10.5
1)∠ABC=∠GBI=y (vert. opp. ∠s)
∠ACB=∠HCE=z (vert. opp. ∠s)
∠BAD=∠ABC+∠ACB (ext. ∠ of △)
x=y+z

2)∠BAC+∠ABC=∠BCE (ext. ∠ of △)
q+2p=2y
y=(2p+q)/2
∠DCE+∠CDE=∠DEF (ext. ∠ of △)
y+q=x
x=(2p+q)/2+q
x+y=(2p+q)/2+q+(2p+q)/2
=2p+q+q
=2p+2q
=2(p+q)

咁大幅 點睇啊
顧下人地感受 =.=

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