See complete Problem 105
Triangle, Angles, Midpoint, Congruence. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Monday, May 19, 2008
Geometry Problem 105
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See complete Problem 105
Triangle, Angles, Midpoint, Congruence. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Angle DBC=180-90+α+β-x-2β-α =90-x-β. Now using the trigonometric version of Ceva's Theorem, we can say that: sin(90-α-β)/sin(x)*sin(β)/sin(β)*sin(α)/sin(90-x-β)=1 or sin(90-α-β)/sin(90-x-β) = sin(x)/sin(α). This equation is clearly true when x=α which makes both sides=1 or equally when x=90-α-β
ReplyDeleteAjit: ajitathle@gmail.com
Ajit:
In particular, if AC=AB, then x=alpha.
ReplyDeleteThen without loss of generality,
we may assume that AC < AB.
Let E be a point on the extended line of AC with AE=AB.
Since the angle DCB=90-x-beta = the angle of DCB, the quadrilateral CDBE is a cyclic
and so alpha =90-beta-x,
that is, x=90-alpha-beta.
September 21, 2009
Bae deok rak
bdr@korea.com
From daechidong,
Topmath academy(02-567-5114)
Seoul, South Korea.
To Antonio: according to the conclusions reached by Joe and Anonymous, the problem 105 has two possible answers for the angle x, meaning that some condition is missing in enunciate. Is that true? What should be changed?
ReplyDeleteDrop a perpendicular from B to meet AD at E & AC at F
ReplyDelete< AFD = 90 - alpha - beta = < BCD
So BCFD is concyclic and
x = < DBF = 90 - alpha - beta
This solution assumes AB not = AC
If they are equal x easily = alpha
Sumith Peiris
Moratuwa
Sri Lanka
AD meet BC in E. Let EF parallel with DC, F holds on BD segment. <BDE=alpha+beta and <DEF=90-alpha-beta. Then FE perpendicular to BD, so DC perpendicular to BD. Then DE=BE=EC. Result AE is bisector and median at the same time, so AB=AC. From symmetry result alpha=x.
ReplyDelete