Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
This entry contributed by Sumith Peiris, Moratuwa, Sri Lanka.
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Tuesday, November 8, 2016
Geometry Problem 1285 Triangle, Quadrilateral, 60, 150, 90 Degree Angles, Congruence, Measurement
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Problem 1285
ReplyDeleteThe sides AB and CD intersects at point K. Let M the medium of KB.Then <KBC=30 or KM=MB=KC.Is triangle ACD=triangle KAD (triangle KAD is equilateral) ,so CD=KB and
AB=KC or CD=2AB=2AD/3.
APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE
How is triangle ACD = KAD?
DeleteSory triangle ACD=KBD.(KD=BD.AC=BD,<ADC=<BKD).
DeletePure Geometry solution
DeleteExtend AB to meet CD at E.
Since AC = BD, the triangles EBD and DCA are similar (SSA)
Therefore BE = CD => EC = AB
Consider the 30-60-90 triangle BEC and
let AB = EC = x
=> BE = CD = 2x and ED = AD = 3x
Hence 2AD=6AB=3CD
Since AC = BD, the triangles EBD and DCA are similar (SSA)
DeleteCould u explain how this is so?
Apostolis - how is KD = BD?
DeleteSumith sory KD=BD.
DeleteNo KD > BD
DeleteExtend AB to meet CD at E and for the equilateral triangle AED
ReplyDeleteLet AD = x
Consider the 30-60-90 triangle BEC
Let CE =y
=> BE = 2y and BC = √3y
=> AB = x-2y and CD = x-y
Applying consine rule to ABC
=> AC^2 = (x-2y)^2+3y^2-2(x-2y)(√3y)cos150
=> AC^2 = x^2+y^2-xy ----------(1)
From Right triangle BCD we have BD^2=3y^2+(x-y)^2
=> BD^2 = x^2+4y^2-2xy ----------(2)
Since (1) = (2)
=> x^2+y^2-xy = x^2+4y^2-2xy
=> xy = 3y^2
=> x = 3y
Therefore AB = y and CD = 2y and AD = 3y
Hence 2AD=6AB=3CD
To Sumit :
ReplyDeleteSince AC = BD, the triangles EBD and DCA are similar (SSA)
Could u explain how this is so?
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In triangle EBD, m(BED) = 60, ED = Length of the side of equlateral triangle
In triangle DCA, m(ADC) = 60, AD = Length of the side of equlateral triangle
and given AC = BD
Hence triangle EBD and DCA are congruent by SSA propertry (two sides ED=AD , AC=BD & Angle m(BED)=m(ADC) = 60)
For congruency by SAS the angle must be the included angle.
DeleteHere this is not the case.
Thank you good to know that SSA alone doesnt work to prove congruence.
DeleteProblem 1285
ReplyDeleteThe sides AB and CD intersects at point K then triangle KAD is equilateral (<ADC=360-60-150-90=60). Draw BE//AD( the point E lies on the CD).Then ABED is isosceles trapezoid (AB=ED).So AC=BD=AE or triamgle ACE is isosceles .Then <AED=<ACK and <DAE=<KAC.But triangles KAC=DAE and AB=ED=KC, KB=CD. But <CBK=30 then CD=KB=2KC or 3CD=6AB.
And AD=AK=AB+BK=AB+2AB=3AB or 2AD=6AB.
APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE