數學 Basic Properties of Circles

2011-01-05 7:24 pm

圖片參考：http://imgcld.yimg.com/8/n/HA00740144/o/701101050034913873436440.jpg

In the figure, ABD and DBC are right-angled triangles, in which <A = <C =90° . N and P are the mid-point of BD and AC respectively.

a. SHow that NP┴AC.

b. If M is the mid-point of BA, show that <ABD = <NPM.

我唔識計B部分=.=
有冇人識計可以計下我=0=?

回答 (1)

用戶25975

2011-01-05 7:45 pm

✔ 最佳答案

b) Join PN, PM and AN, then draw a circle with NB = ND being radii.

The circle drawn will pass through both A and C since ∠BAD = ∠BCD = 90 (∠ in semi-circle)

Therefore ABCD is a cyclic quad.

Also since M is mid-pt. of AB, NM is perp. to AB and therefore APNM is a cyclic quad with the reason ∠AMN + ∠APN = 90 + 90 = 180

So we have:

∠MAN = ∠NPM (∠s in the same segment)

Also, △NAB is an isos. △ with NA = NB (radii of the circle)

Hence ∠MAN = ∠ABD (Base ∠s, isos. △)

Finally we have ∠NPM = ∠MAN = ∠ABD

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