ABCD is a rectangle. and a straight line APQ cuts BC in P & DC extended in Q. Locate the point P so that the sum of the areas of the two triangles ABP&CPO may be a minimum.
ABCD is a rectangle. and a straight line APQ cuts BC in P & DC extended in Q. Locate the point P so that the sum of the areas of the two triangles ABP&CPO may be a minimum.
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如图,已知正方形ABCD的边CD上任意一点E.延长BC到F,使CF=CD.设BE与DF相交于G,求证:BG⊥DF.
于四边形之内,取一点不在两对角线之交点之上者,试证明从此点至各顶点之距离之和大于两对角线之和.
PQRS为平面四边形,QR=1,∠PQR= ∠QRS= 70°,∠PQS=15°,∠PRS= 40°.若∠RPS=θ.PQ=α,PS=β,则4αβsinθ属于下列哪个区间【 】
⊙O 的半径是 a,ABCD 是它的内接四边形,∠A =75°,∠B = 120°,AB = BC,求四边形各边长.
如图,∠ABC=90°,AC=BC,AD⊥CE,BE⊥CE,垂足分别是D,E.已知AD=8,BE=3,则DE=______.
已知直线 y = kx + b (k > 0) 与圆 x2 + y2 = 1 和圆 (x − 4)2 + y2 = 1 均相切, 则 k = _______, b = _______.
如图,AB是半圆的直径,C是半圆上一点,直线MN切半圆于C点,AM⊥MN于M点,BN⊥MN于N点,CD⊥AB于D点 . 求证:(1) CD=CM=CN;(2) CD2=AM•BN.
已知△ABC三内角的大小成等差数列,tanAtanC=2+,求角A,B,C的大小;又知顶点C的对边c上的高等于4,求三角形各边a,b,c的长.(提示:必要时可验证(1+)2=4+2)