2021-2022 school year liwan Eighth grade first semester math final question

2022-06-24 0 By

As shown in the figure, two triangles with common vertices △ABC, △AB ‘C’, if AB=AB ‘, AC=AC ‘and ∠BAC+∠B’ AC ‘=180°, we say △ABC and △AB’ C ‘are complementary triangles to each other.(1) As shown in the figure, △ABC is an isosceles triangle, △ABE and △ACD are isosceles right triangles, which are connected with DE;Verify: △ABC and △ADE are topcomplement triangles.Under the condition of (1), BE intersects CD at point F, connects AF and extends the intersection BC at point G.Determine the quantitative relationship between DE and AG and prove your conclusion.In the quadrilateral ABCD, ∠B=40°, ∠C=50°.Is there a point P in the plane where △PAD and △PBC are topcomplement triangles? If so, draw a graph and prove it.If not, please explain the reason.First (figure 3) ask answer: (1) the first (1) as shown in figure as auxiliary line ask delta is isosceles triangle ABC = > | AB | = | AC |.Delta ABE, delta ACD is isosceles right triangle = < = > < DAC BAE = 90 ° | AB | = | AE | | AC | = | | AD.Suppose Angle BAC=x, Angle DBE= EAC=90°-x.| AB | = | AE AC | | = | = | | AD.DAE+ BAC=90°+90°-x+ X =180°△ABC and △ADE complement each other.(2) do auxiliary line, extend the DA to B 'point | DA | = "| | AB, connection B' E, G 'for B & # 39;E midpoint, connecting AG prime.DAE+ BAC=180° ∠DAE+∠B 'AE=180° ∠BAC= ∠B' AEAC | | AB | = | AE | | = | | = | AB AD '| = > delta BAC ≌ delta B’ AE.| | = | AD AE | = < < DAC EAB = Rt < | AB | = | AC | = > delta DAC ≌ delta ADC = the EAB < < AEB < 1 = 2 < < 3 = < 4.AG is a perpendicular bisector of line segment DE.< BAG = < the CAG.AC | | AB | = | | | = | | AG AG.Delta BAG ≌ delta the CAG.AG is the perpendicular bisector of line segment BC.Similarly, AG prime is the perpendicular bisector of segment B 'E.|AG|=|AG'|.Quadrilateral AHEG & # 39;Is a rectangle.|AG'| = | HE | | | = 0.5 * DE.| DE | | | AG = 2.(3) if the point P is the delta PAD to delta PBC are top triangle, have | PA | = | PB | | | | = | PC PD.The point P lies on the perpendicular of line segment AB and line segment CD.For the auxiliary line, PG is neutral perpendicular AB, PH is neutral perpendicular CD.PG intersects PH at the point P.Connect PA, PB, PC, and PD.Pictured above (3) | PA | = | | in PB, | | | = | PC PD.Suppose APB is equal to x and CPD is equal to y.△APB is an isosceles triangle.∠PAB= (180°-x) /2=90°-x/2.And similarly, PDC is equal to 90 degrees minus y over 2.< AEB = 40 ° 180 ° - - < PAB = 50 ° + x / 2.In the same way < DFC = 40 ° + y / 2.< EPF = 180 ° - < AEB - < DFC = 90 ° + 2 + x/y / 2.< < the BPC + EPF = x + 90 ° - 2 + 2 x/y/y + 90 ° - 2 x/y / 2 = 180 ° delta PAD to delta PBC are top triangle.Point P exists.