本帖最後由 lex 於 14-6-8 20:47 編輯
回覆 lm251099 的帖子
Maybe you(and your beloved child) have already figured out a way to solve these maths problems, but here is my two cents ----------------- 27A) In Triangle ABC, EG//BC[given] AG=GC[given] Therefore, EB=AE=3cm [mid.pt. thm]
27B) In Triangle ABC EG//BC[given] AG=GC[given] AE=EB[proved] Therefore, EG=0.5(BC)[mid. pt. thm]
Because HC=BH [given] and BH+HC=BC
Therefore,HC=EG=3cm
27C) In Triangle FDE, HC//ED, EH=HF[given] Therefore, DC=CF[mid. pt. thm] and HC=0.5(ED)[mid. pt. thm]
Therefore, ED=2 HC=2(3)=6cm
In Polygon BCDE, BC//ED BC=2(EG)=6cm=HC Therefore BCDE is a parallelogram [hmm....I cant recall the exact reason for this one....But i am positive that it is correctly proved in this way]
--------------------------------- 28A) In Parallelogram ABDE, AB=ED [prop of //gram] In Parallelogram BCDE, ED=BC [prop of //gram] Therefore, AB=ED=BC
28B) In Parallelogram BCDE, DM=MB,EM=MC [property of //gram]
In Triangle DBC, DP=PC[given] DM=MB[proved] Therefore, MP=0.5(BC) [mid.pt. thm]
Because AB=BC[proved]MP=0.5(AB)
28C) Using the result of 28B, MP//BC[mid.pt. thm] Because PM produced at Q and R, Therefore, PMQR is a st.line
and because ABC is a st. line, Therefore RP//AC
In Triangle EBD, DM=MB[proved] QM//ED[proved] Therefore QM=0.5(ED) [mid. pt. thm] Because ED=AB [prop. of //gram] Therefore QM=0.5(AB)
In Triangle EAB, EQ=QB[prop. of //gram] RQ//AB[proved] Therefore RQ=0.5(AB)[mid.pt. thm]
Because MP=0.5(AB)[proved] Therefore RQ=QM=MP
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