Plzzzzz help me with these sums ..i hav done many sums to revise for my exam tmrrw..but stuck with these. help

Question

AB {ll (parallel to)} DE. Prove‘angle’ABC + ‘angle’BCD =180 + ‘angle’CDE
http://img102.imageshack.us/img102/1672/newbitmapimagemi5.png

Prove:AB { ll (parallel to)} EF
http://img134.imageshack.us/img134/1913/newbitmapimagevy2.png

Find value of x.
http://img412.imageshack.us/img412/774/newbitmapimageel8.png

‘angle’DFP, ‘angle’ EDQ & ‘angle’ FER are exterior angles of ‘triangle’ DEF. Prove: ‘angle’DFP + ‘angle’ EDQ + ‘angle’ FER = 360
http://img266.imageshack.us/img266/3655/newbitmapimagepk8.png

BD & CD are angle bisectors of ‘angle’ ABC &‘angle’ACE respectively.Prove ‘angle’ BDC=½ ‘angle’BAC
http://img86.imageshack.us/img86/8948/newbitmapimagenm9.png

PT (perpendicular to) QR & PS bisects ‘angle’QPR. If ‘angle’ Q = 65 & ‘angle’ R = 33. Find ‘angle’ TPS
http://img503.imageshack.us/img503/4818/newbitmapimage2vm6.png

ABCD & BPQ are lines.BP=BC & DQ {ll(parallel to)} CP.Prove CP=CD & DP bisects ‘angle’CDQ
http://img249.imageshack.us/img249/498/newbitmapimage2ax1.png

Answer 1

AB {ll (parallel to)} DE. Prove‘angle’ABC + ‘angle’BCD =180 + ‘angle’CDE
http://img102.imageshack.us/img102/1672/...
~~~Not hard.
Draw a line // to AB & DE and apply "alt. angles, CF//AB".
`
Prove:AB { ll (parallel to)} EF
http://img134.imageshack.us/img134/1913/...
~~~1st, prove AB//CD & EF//CD, then use "transitive property of // lines" to prove AB//EF.
`
Find value of x.
http://img412.imageshack.us/img412/774/n...
~~~Apply ext. angle of triangle twice.
`
'angle’DFP, ‘angle’ EDQ & ‘angle’ FER are exterior angles of ‘triangle’ DEF. Prove: ‘angle’DFP + ‘angle’ EDQ + ‘angle’ FER = 360
http://img266.imageshack.us/img266/3655/...
~~~Just state: angle’DFP + ‘angle’ EDQ + ‘angle’ FER = 360
(sum of ext. angles of polygon)
Well, a triangle is a kind of polygon, right?
`
BD & CD are angle bisectors of ‘angle’ ABC &‘angle’ACE respectively.Prove ‘angle’ BDC=½ ‘angle’BAC
~~~This simply needs those reasons involving triangles.
`
PT (perpendicular to) QR & PS bisects ‘angle’QPR. If ‘angle’ Q = 65 & ‘angle’ R = 33. Find ‘angle’ TPS
http://img503.imageshack.us/img503/4818/...
~~~The picture does not match the question!
`
ABCD & BPQ are lines.BP=BC & DQ {ll(parallel to)} CP.Prove CP=CD & DP bisects ‘angle’CDQ
http://img249.imageshack.us/img249/498/n...
~~~Must use (base angles, isos triangle) and (sides opp. eq. angles) multiple times.

Answer 2

Extend AB and DC so they meet at F
Then angle BFC = angle CDE [alt int angles are =
angle FBC = 180 - angle ABC [ supplementary angles]
angle BCD = angle FBC+angle BFC [ext angle = sum int]
Thus angleBCD =180 - angleABC+angleCDE{Substitution}
Hence angle BCD+anleABC= 180+abgle CDE[algebra]

66 = 30 +66, so AB||CD [alt int angles =]
30+50 = 180,so EF || CD [same side int angles supplements]
Thus AB ||EF { both lines || CD}

abgle DBE = 34+30=74 [ext angle = sum of opp int angles]
angle x = 74+45 =129 [ext angle = sum of opp int angles]

angle FER+ angle FED = 180 [supplementary angles]
angle EDQ + angle EDF = 180 {same reason}
angle DFP +angle DFR = 180 [same]
Thus angle FER +angle EDQ +angle DFP+angle FED+
angle EDF + angle DFR = 540 [addition axiom]
But angle FED +angle EDF + angle DFR =180 [triang sum]
Thus angleFER+angleEDQ+angleDFP = 540-180 =360
{Substitution and Subtraction axioms}

Sorry, ran out of time.