Please use the method of mid-pt theorem or intercept theorem.
1) Q, R is the mid-points of the sides AC, AB of △ABC; AD is the perpendicular from A to BC. Prove that ∠RDQ = ∠BAC.
2) ABCD is a parallelogram and HK is a straight line outside the paralellogram. AP, BQ, CR, DS are the perpendicualrs from A, B, C, D to HK. Prove that AP + CR = BQ + DS (Hint: Let AC cut BD at O; draw ON perpendicular to HK.)
3) BR, CQ are the bisectors of ∠B, ∠C of △ABC; AH, AK are the perpendiculars from A to BP, CQ; prove that KH is parallel to BC. (Produce AH, AK to cut BC at X,Y)
THX
Maths (Geometry) (15x2)
2009-04-09 7:55 pm
回答 (1)
2009-04-10 1:17 am
✔ 最佳答案
Q1.Since AD is perpendicular to BC, so angle ADC is a rt. angle and AQ = QC, so Q is the centre of circle with AC as diameter, so DQ = QC = AQ.
Similarly, DR = BR = AR.
For triangle ARQ and triangle QRD
AR = RD (proved)
AQ = DQ (proved)
RD = RD (common)
so they are congruent (SSS)
so angle RAQ = angle ABC = angle RDQ.
Q2.
Let BS cut ON at X.
Since BO = OD and ON //DS//BQ, by mid-point theorem,
DS = 2OX.
and BQ = 2XN.
Adding together, we get DS + BQ = 2OX + 2XN = 2(OX + XN) = 2ON.
Similarly, AO = OC, AP//ON//CR, by mid-point theorem,
AP + CR = 2ON
therefore, AP + CR = DS + BQ.
Q3.
For triangle ABH and triangle HBX
angle ABH = angle HBX ( BR is the angle bisector)
BH = BH (common)
angle AHB = angle BHX = 90 degree ( AH perpendicular to BR)
therefore, the 2 triangles are congruent (ASA)
so AH = HX.
Similarly, triangle AKC congruent triangle CKY (ASA)
so AK = KY.
that is K is the mid-point of AY and H is the mid-point of AX, by mid-point theorem, KH//XY or KH//BC.
2009-04-09 17:49:36 補充：
For Q1, if insisting on using mid-point theorem, then let AD intersect RQ at X, triangle AXQ congruent triangle QXD (AX = XD by mid-point theorem), and triangle ARX congruent triangle XRD, so angle RAQ = angle RDQ.
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