Maths (Geometry) (hard) (20x2)

2009-03-28 5:54 am

Please solve this problem by using some method in deductive geometry or you can use other methods (e.g. substitution, adding lines). But please dont use trigo. method. THX!!!!

1) ABCD is a square; E is a point on AC such that AE = EB; the line through E perpendicular to AC cuts BC, DC at F, G. Prove that ∠FAG = 45.
2) In △ABC, ∠C is a right angle; ABPQ is a square outside △ABC; PN is the perpendicular from P to AC. Prove that PN = CA + CB.

It's better if someone can draw diagrams and give detail explanation. THX A LOT !!!

更新1:

Sorry!!!!!! AE=AB

更新2:

Hey!!! Please answer (1)!!!!!! (AE=AB)

回答 (2)

用戶10012

2009-03-28 3:33 pm

✔ 最佳答案

Q1. Really AE = EB ?

Q2.
Let M be a point on PN such that BM is perpendicular to PN, therefore MN = BC......................(1)
For triangle PBM and triangle ABC
PB = AB ( side of a square)
Angle PMB = Angle C = rt. angle
Angle MPB = Angle A (this can be proved easily by similar triangle, see remark below).
Therefore the 2 triangles are congruent (AAS).
Therefore PM = AC..................(2)
From (1) + (2) we get
PM + MN = CA + CB = PN.
Remark:
Let AB intersects PN at H.
For triangle HPB and triangle AHN
Angle PBC = rt angle (property of square) = angle HNA (PN perpendicular to AC).
Angle PHB = angle AHN ( vert. opposite angles)
so triangle HPB similar to triangle AHN (AAA),
therefore angle MPB = Angle A.

2009-03-28 07:37:58 補充：
Correction : 3rd line under remark should be angle PBA, not angle PBC.

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用戶9170

2009-03-28 6:18 am

Are you sure no writing mistake in the qu? AE = EB should be AB = EB or AE = AB?

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