See complete Problem 59
Right and Equilateral Triangles, Midpoints. Level: High School, SAT Prep, College geometry
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Monday, May 19, 2008
Elearn Geometry Problem 59
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We let BA & BC be the x-axis & y-axis resply. and use standard nomenclature for Yr. ABC by calling EF=p and AD=q. We, therefore, need to prove that 4p^2=q^2. We've A:(c,0),B:(0,0),C;(0,a) and angle ABD=30 deg. Hence, D:(V3a/2,a/2),
ReplyDeleteF:(V3a/4,a/4) and E:(c/2,a/2) since BC=BD=a.
4p^2 =4[(V3a/4-c/2)^2+(a/4-a/2)^2
= (V3a/2-c)^2+(-a/2)^2
=(V3a/2-c)^2+(a/2)^2 -----------(1)
while q^2 = (V3a/2-c)^2 + (a/2)^2 ---(2)
By (1) & (2), it is evident that 4q^2 = q^2 or 4EF^2=AD^2 or EF = AD/2
Ajit: ajitathle@gmail.com
http://geometri-problemleri.blogspot.com/2009/06/problem-27-ve-cozumu.html
ReplyDeletehttp://img402.imageshack.us/img402/4481/problem59.png
ReplyDeleteLet G is the midpoint of CD
Connect BE & EG ( see sketch)
Since E is the midpoint of AC => ∆BEC is isosceles and BE=EC, ∠EBC=∠ECB= α
∠FBE=∠ECG=60- α
Triangle BFE congruence to ∆CGE …. ( Case SAS)
So EF=EG= x
E,G are midpoints of AC & CD => EG= a/2= x
G midpoint CD, H midpoint BC
ReplyDeleteEH // AB ; EH is perpendicular bisector of BC passes D ; FG //BC ; DGEF is a kite; FG=EF; FG=1/2 AD ;EF=1/2 AD
Draw circumcircle Θ of center E and diameter AC to ΔABC. Extend BE to intersect Θ in G.
ReplyDeleteSince AC and BG are diameters of circle Θ, ΔΑΒC and ΔGCB are rectangle respectively in B and C, and since <CGB = <CAB, then ΔΑΒC and ΔGCB are similar with same hypotenuse lengths. Therefore GC = AB (1).
By construction, <ABD = <GCD = 90°-60° = 30° (2).
Since ΔBCD is equilateral, then BC = CD (3).
From (1) , (2) and (3) above, ΔABD and ΔGCD are SAS with sides of equal lengths therefore AD = DG = a (4).
In ΔBDG, F middle of BD (given) and E middle of BG (by construction) therefore FE = DG/2 = x (5).
From (4) and (5) above, a = x/2, QED.
Let K,L be midpoints of BC, AB; D-E-K are collinear, BLEK a rectangle, thus EL is projection of BD onto the line EL, and the midpoint of BD lies onto the perpendicular bisector of DE, thus EF=LF, but LF is midline in triangle BAD, done.
ReplyDeleteConsider Triangles EBF & DCB.
ReplyDeleteFB = DC/2 and BE = AC/2. Moreover the included angles < FBE = < ACD since < FBC = < ECB
Hence the 2 triangles are similar and x = a/2
Sumith Peiris
Moratuwa
Sri Lanka
No diagram is shown
ReplyDelete