Q1- ABCD is a rectangle in which BC=2AB.A point E lies on ray CD such that CE=2BC.Prove that BE Perpendicular AC.
Q2-In a Triangle ABC, AD is a median .X is a point on AD such that AX:XD=2:3.Ray BX intersect AC in Y.Prove that BX=4XY.
2006-12-14 21:48:16 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
AB has to be 2BC !
anyways I think it can be done with triangles that are formed inside the rectangle. Cheers
2006-12-14 22:03:53 · answer #1 · answered by Tim 3 · 1⤊ 0⤋
let BE intersects of eatch AC,AD at G,F
AB paralle CE ,then triangle ABGis similar to triangle CEG(given)AB=1/2BC=1/2CE then AB:CE=1:4alsoBG:GE=1/4
Triangle ABFsimilar to BEF and since AB=1/3DE theFE=1/3
from ratios we deduce that BG:GF:FE=4:1:15
Triangle BCE is right at C,let BC=2L ,then CE=4L
We deduce BE=2(times square root of 5)L,thenEF=15*2(TIMES ROOT OF 5)/20L=3(ROOT 5)/2L
Also we deduceEG=16*2(ROOT 5)/20=8(ROOT5)/5L
Since ED:EG=3L divides 8(root5)/5L=115:8(ROOT5)
AlsoEF:EC=the same ratio andangleE IS COMMON In two triangles EDF,EGC THEN they similar the result is angles EDF,EGC ARE EQUAL
THEN AC PERPENDICULAR TO BE
2006-12-15 08:32:41 · answer #2 · answered by eissa 3 · 0⤊ 0⤋
Q1.
Slope of AC = AB/BC = 1/2
Slope of BE = CE/BC = -4/2 = -2.
Since the slopes are negative reciprocals AC is perpendicular with BE.
Q2. Please clarify. I can't figure out the diagram.
2006-12-15 06:26:00 · answer #3 · answered by Northstar 7 · 1⤊ 0⤋
unfortunately a paper can not be used here..but this is a hint to the to the answer::::
there should be a rectangle and two triangles, each at one end of the rectangle..its gonna look like a trapezium..
2006-12-15 06:18:53 · answer #4 · answered by George 2 · 1⤊ 0⤋
this is a question which should be done in paper.
u can ask ur teacher.
2006-12-15 05:58:27 · answer #5 · answered by avanthi 2 · 0⤊ 2⤋