Online Geometry theorems, problems, solutions, and related topics.
See complete Problem 105Triangle, Angles, Midpoint, Congruence. Level: High School, SAT Prep, College geometryPost your solutions or ideas in the comments.
This Problem has two solutions for xx=Alfax=90-Alfa-Beta
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-βAjit: email@example.comAjit:
In particular, if AC=AB, then x=alpha.Then 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, 2009Bae deok firstname.lastname@example.orgFrom 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?