中考几何动态真题
(1)
如图1，∵BA⊥AM，MN⊥AP，∴bam = anm = 90∴∠paq+∠man =∠man+∠amn = 90∞。
mn⊥ac,∴∠pqa=∠anm=90 ∴aq=mn,∴△aqp≌△mna∵an=pq
AM=AP，∴∠AMB =∠APM∠APM =∠BPC∠BPC+∠PBC = 90，∠AMB+∠abm = 90∞。=个人电脑
(2)
(1)方法一:如图1，∵BA⊥AM，MN⊥AP，∴ BAM = ANM = 90 ∴∠ PAQ+∠ MAN+。
mn⊥ac,∴∠pqa=∠anm=90 ∴aq=mn,∴△aqp≌△mna∵an=pq
AM=AP，∴∠AMB =∠APM∠APM =∠BPC∠BPC+∠PBC = 90，∠AMB+∠abm = 90∞。方法二:如图1，∵BA⊥AM，MN⊥AC，∴∠ BAM = ANM = 90 ∴∠ PAQ+∠ MAN = ∠ AMN = 90。∴∠APM =∠amp∠aqp+∠bam = 180，∴pq∨ma∴∠qpb =∠amp ∴.
PC=3，∴来自(1)，PC=AN=3∴AP=NC=5.
AC=8，∴am = AP = 5∴AQ = Mn = = 4≈paq =∠amn∠ACB =∠anm = 90∴∠ABC =∞。那么CF=3k∴=，ne = K .设n为NTFE，CF为t，那么四边形ntfe为平行四边形ne = tf = k，∴CT = cf∴TF = 3k÷k = k÷ef .∴tan∠NTC==2，∴CT=k=，∴k=，∴CK=2×=3，bk = BC∶CK = 3∶PKC+∶dkc。Tan∠ABC=，∴设GK=4n，那么BG=3n，GD=4n∴BK=5n=3，∴n=，∴BD=4n+3n=7n=∴AB==10.
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