线段的调和分割在证明两角相等中的应用

设PC、PD分别为△PAB的∠APB的内、外角平分线,由三角形内、外角平分线性质,可得AC／CB=PA／PB=AD／DB,更一般地,若两点C、D内分与外分同一线段AB成同一比值,即AC／CB=AD／DB,则称点C、D调和分割线段AB．显然,当C、D调和分割AB时,也有A、B调和分割CD：CA／AD=CB／BD,如图1,其中PA,PB,PC、PD也称为调和线束．     

从中考题看圆中辅助线的添置法

一、作半径造圆心角,与同弧上的圆周角相联系例1如图1,O的直径AB的延长线与弦CD的延长线交于P,E为O上一点,AE＝AC,DE交AB于F．求证：PF·PO＝PA·PB．（1997年河北省中考题）分析PA·PB＝PC·PD,欲证结论成立,只须证PF·PO＝PC·PD,即只须证PF／PD＝PC／PO．为此,只须证△PDF△POC。／P公用,…只须证上FDP一工COP．连结CO,”．’AE＝AC,…／l＝／2．用等角的补角相等获证．二、过圆心作弦的垂线,以便应用垂径定理例2如图2,AB是①O的弦,P是AB上一点,AB＝10cm,PA＝4cm,OP＝scm．求①O的半…     

简证一道联赛题

题如图1,已知等腰三角形△ABC中,AB=AC,∠C的平分线与AB边交于点P,M为△ABC的内切圆⊙I工与BC边的切点,作MD／／AC,交⊙I于点D．证明：PD是⊙I的切线．     

一道初中数学联赛平几题证法的探究

2010年全国初中数学联赛平面几何题为： 如图1,已知等腰三角形△ABC中,AB=AC,∠C的平分线与AB边交于点P,M为△ABC的内切圆⊙I与BC边的切点,作MD／／AC,交⊙I于点D,证明：PD是⊙I的切线．     

Rotenone induces more serious learning and memory impairment than α-synuclein A30P does in Drosophila

Parkinson’s disease (PD) is a heterogenous disease caused by multifactorial etiology. PD is characterized by the loss of dopaminergic (DA) neurons in the substantia nigra and the accumulation of Lewy bodies. In this study, two Drosophila PD models by exposing Drosophila to rotenone (sporadic PD models) are proposed, and the human α-synuclein A30P protein (family PD models) in Drosophila is expressed respectively. Both models recapitulated the main human PD symptoms including the loss of dopaminergic neurons...     

矩阵广义正定性的充要条件

在二阶及上 (下 )三角矩阵的情况下证明了A∈PD 的充要条件 ,并由此说明了PI,PD,PS 三者之间的关系 ,若A∈R2× 2 ,有PI PD ={一切主子式大于零的矩阵 } PS;若A为上三角阵 ,则PD ={一切主子式大于零的矩阵 } .     

从一道几何证明题谈面积法

正如图1,已知在△ABC中,AB=AC,P是BC上任一点,PD⊥AB于D,PE⊥AC于E,CF⊥AB于F.求证:CF=PD+PE.对于该题,一般同学会想到截长法与补短法.如图2,过点P作PM⊥CF于M,则四边形PMFD是矩形,则PD=FM.易证△PCM≌△CPE,则CM=PE.于是CF=FM+CM=PD+PE.这种方法叫做截长法.如图3,过点C作CN⊥DP交DP的延长线于点N,则四边形NCFD是矩形,则CF=DN.易证△CPN≌△CPE,则PN=PE.于是CF=DN=PD+PN=PD+PE.这种方法叫做补     

Extraction of Partial Discharge Acoustic Signal by Wavelet Transform with Teager＇s Energy Operator

To develop a measurement system for monitoring partial discharge （PD） without the effect of external interferences,an algorithm of PD signal extraction based on wavelet transform with Teager＇s energy operators was presented. Acoustic signal generated by PD was selected to remove excessive interfering signals and electromagnetic interferences. Acoustic signals were collected and decomposed into I0 levels by wavelet transform into approximation and detail components. “Daubechies 25” was proved to be the most suitable mother wavelet for the extraction of PD acoustic signals. Compared with conventional wavelet denoising method, Teager＇s energy operators were adopted to the PD signal reconstruction and the signal to noise ratio was in creased by 20%-25% inthe experiment,without lost in energy and pulse amplitude.     

高考全国理科卷2第（20）题别解

题目如图1,四棱锥P-ABCD中,底面ABCD为矩形,PD⊥底面ABCD,AD=PD,E,F分别为CD、PB的中点.     

不同播期夏玉米籽粒灌浆特性及与主要气象因子的关系

试验选用玉米杂交种郑单958(ZD958)和隆平206(LP206)为材料,分别于6月9日(PDⅠ)、6月17日(PDⅡ)和6月26日(PDⅢ)等3个时期播种,比较研究了玉米产量、籽粒灌浆特性及籽粒生长期间主要气象因子的变化。结果表明,中期播种的玉米籽粒产量最高,早播处理次之,晚播处理产量最低。PDⅡ籽粒启动势和灌浆持续期显著高于PDⅠ,同时PDⅡ籽粒平均灌浆速率和最大灌浆速率显著高于PDⅢ。可见,播期主要通过籽粒灌浆启动时间的早晚、灌浆速率的高低以及灌浆持续时间的长短影响籽粒生长。籽粒灌浆期间有效积温与平均灌浆速率呈显著正相关,而平均日温差与启动势呈显著正相关。说明玉米籽粒生长期间有效积温不足是晚播玉米减产的主要因素,而早播主要是灌浆期间平均日温差较低造成的。沿淮地区玉米生长的适宜播期在6月17日左右。     

一题多变 触类旁通

题目已知:如图1,AM是△ABC中BC边上的中线,P是AM上任意一点,过点P作DE∥BC,交AB、AC分别于D、E. 求证:PD=PE. 证明:∵DE∥BC, ∴(PD)/(BM)=(AP)/(AM),(PE)/(MC)=(AP)/(AM),∴ (PD)/(BM)=(PE)/(MC), ∵BM=MC,∴PD=PE. 变式一已知:如图2,AM是△ABC中BC边上的中线,P是AM上     

2005年高考数学（理）第20题别解

题目 如图1，四棱锥P—ABCD中，底面ABCD为矩形，PD┴底面ABCD，AD=PD，E、F分另9为CD，BP的中点．     

2005全国高考立体几何第20题(Ⅱ)别解

2005全国高考(Ⅱ)卷20题如图,四棱锥P-ABCD中,底面ABCD为矩形,PD⊥底面ABCD,AD=PD,E、F分别为CD、PB的中点.     

物理显影记录系统

简述了物理显影记录系统的原理和特性 ,并讨论了PD -D系统和PD -R系统 ,彩色物理显影以及物理显影的应用等 ,以期对物理显影有着更深入的理解     

Extraction of Partial Discharge Acoustic Signal by Wavelet Transform with Teager′s Energy Operator

A big operational problem of high voltage (HV)power equipments is the degradation of the insulation,to which partial discharge (PD) is a major contribu-tion. Currently, the main collectingtechniques for PDsignals include the electrical pulse associated withPD, the electromagnetic waves detection method andthe acoustic detection method. PD generates acousticwaves, which can be used to detect the intensity andlocation of the PD. Compared with the detectionmethods, the acoustic methods have a…     

Investigating the role of ^99mTc-TRODAT-1 SPECT imaging in idiopathic Parkinson＇s disease

Objective: To investigate the role of 99mTc-TRODAT-1 SPECT in diagnosis and assessing severity of idiopathic Parkinson's disease (PD). Methods: Thirty-eight patients with primary, tentative diagnosis of PD and eighteen age-matched normal controls were studied with 99mTc-TRODAT-1 SPECT imaging. The regions of interests (ROIs) were drawn manually on cerebellum (CB), occipital cortex (OC) and three transverse plane slice-views of striatums, the semiquantitative BG (background)/[(OC CB)/2] were then calculated. Results: A lower uptake of 99mTc-TRODAT-Ⅰ in striatums were displayed in thirty-six out of thirty-eight PD patients by visual inspection, compared to controls. In twenty-four PD cases with HYS (Hoehn and Yahr scale) stage I, a greater loss of DAT uptake was found in striatum and its subregions contralateral striatum to the affected limbs than in the same regions of the controls, although the striatal uptake was bilaterally reduced. Using Spearman correlation analysis showed that the reduction of the uptake ratios significantly correlated with the UPDRS in striatum and all its subregions in the PD group (P＜0.05), a similar change was also found in the putamen by using the rating scale of Hoehn and Yahr (P＜0.05).However, analysis of variance (ANOVA) did not show any relationship between the decreasing uptake of 99mTc-TRODAT-1 and increasing severity of PD patients, although the specific uptake of 99mTc-TRODAT-Ⅰ was continuously decreased in the striatum by visual inspection with the progress of PD from HYS stage Ⅰ to Ⅲ. Conclusion: 99mTc-TRODAT- 1 SPECT imaging may serve as a useful method for improving the correct diagnosis of PD. In assessing the role of99mTc-TRODAT- 1 SPECT in disease severity of PD, UPDRS can offer a comprehensive index, although the Hoehn and Yahr assessment may be available in part.     

从一道几何证明题谈面积法

如图1，已知在AABC中，AB=AC，P是BC上任一点，PD⊥AB于D，PE⊥AC于E，CF⊥AB于只求证：CF=PD＋PE．     

切割线定理的应用

切割线定理及推论是圆中的重要定理,应用十分广泛,切割线定理有用,但怎样用是很有讲究的. 一、根据条件直接用例1 已知P是两圆公共弦BA的延长线上的任意一点,PC和PD为从点P到两圆所引的切线,C、D为切点.求证:PC=PD. 证明:∵PC、PD是圆     

立几中的割补法

1．求二面角例1如图1,在四棱锥P—ABCD中,PD 平面ABCD,底面ABCD为矩形,若AB=4,BC=5,PD=3,求面PAB与面PCD所成二面角的大小．     

矩阵广义正定性的充要条件

采用待定法证明了二阶及上 (下 )三角矩阵A∈PD 的充要条件 ,把广义正定性问题转化为实对称矩阵A∈PD 的正定性问题 ,与文 [1]比较降低了证明难度 ,大大简化了证明过程。