中三級數學題(2)

2012-10-28 12:15 am

1.在圖中，ABCD是一個邊長為3的正方形。E是DC上的一點，使DE=1。AE的垂直平分線分別與AD﹑AE和BC 相交於P﹑M和Q。求PM/MQ。

圖片參考：http://imgcld.yimg.com/8/n/HA07015327/o/20121027161132.jpg

2.在圖中，ABCD是一個正方形。M和N分別是AB和BC上的點，使BM=BN。在MC 上取一點P，使BP ⊥ MC 。證明PD ⊥ PN。

圖片參考：http://imgcld.yimg.com/8/n/HA07015327/o/20121027161351.jpg

更新1:

唔明: PM : (√10/2) = DE : DA = 1 : 3 PM = √10/6 為何 ∠BCM = ∠PBM

更新2:

∠BCM = ∠PBM 已明白 why △NBP ~ △DCP

更新3:

why: PM : (√10/2) = DE : DA = 1 : 3 PM = √10/6

更新4:

BP/BM = PC/BC because △BPM ~ △CPB but why △BPM ~ △CPB，which are the 3 angles of the △BPM ~ △CPB of reasons？？please explain it﹗﹗ Also can I say BP/BM = PC/BC because △BPM ~ △CBM？？？ please help me to explain it﹗thanks﹗

回答 (3)

用戶2053

2012-10-28 8:16 pm

✔ 最佳答案

1)
AM = (1/2)AE = (1/2)√(3² + 1²) = √10/2
PM : (√10/2) = DE : DA = 1 : 3
PM = √10/6
PQ = AE = √10
MQ = PQ - PM = √10 - √10/6 = (5/6)√10
∴ PM / MQ = √10/6 / [(5/6)√10] = 1/5
2)
BP/BN = BP/BM = PC/BC = PC/CD ....(1)
∠DCP = 90° - ∠BCM
及
∠NBP = 90° - ∠PBM
因∠BCM = ∠PBM
故 ∠DCP = ∠NBP ....(2)
(1) & (2) :△NBP ~ △DCP
得 ∠NPB = ∠DPC
而 ∠CPN + ∠NPB = 90°
即 ∠CPN + ∠DPC = 90°
⇒ ∠DPN = 90°
⇒ PD丄PN

2012-10-28 12:55:01 補充：
why:
PM : (√10/2) = DE : DA = 1 : 3
PM = √10/6

Because :
△ APM ~ △AED

2012-10-28 12:58:10 補充：
△NBP ~ △DCP

Because BP/BN = PC/CD ( proved in (1) )
and ∠DCP = ∠NBP
(SAS)

2012-10-29 23:41:41 補充：
You cann't just say BP/BM = PC/BC because △BPM ~ △CBM.

First prove △BPM ~ △CBM :
∠PMB =∠BMP (Common) ,
∠BPM =∠CBM = 90°
(A.A.)
So we have ∠PBM =∠PCB
On the other hand , ∠BPM =∠CPB = 90°
So △BPM ~ △CPB (A.A.)

Then you can say it.

👍 0 👎 0

DL081797

2012-10-30 1:41 am

BP/BM = PC/BC because △BPM ~ △CPB
but why △BPM ~ △CPB，which are the 3 angles of the △BPM ~ △CPB of reasons？？please explain it﹗﹗
Also can I say BP/BM = PC/BC because △BPM ~ △CBM？？？
please help me to explain it﹗thanks﹗

👍 0 👎 0

小銘

2012-10-29 5:27 am

寫的好阿
只是要看粉久@@

👍 0 👎 0

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