Triangle problem
In the triangle ABC, find x. (using elementary geometry only). Note: The triangle is not drawn to scale. If you want you can construct some more lines.
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Vasavi GS
8 years, 2 months ago
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17 votes
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A complete discussion of this well-known problem can be found here, so I find it unnecessary to furnish a detailed solution: http://www.cut-the-knot.org/triangle/80-80-20/index.shtml
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I get two down votes for linking to a complete and correct solution that satisfies the criterion that nothing more than elementary geometry should be used, while several others post incorrect answers with no explanation whatsoever? Are others really so lazy that they cannot be bothered to follow a link? Fine. Here is one method of solution, using the same notation as in the original post:
First, locate point F on EB such that ∠FAB=50∘. Then we first show that ∠FDB=30∘. To see this, locate G on EB such that draw DG∣∣AB, and it easily follows that the intersection of GA and DB at P forms two equilateral triangles △DGP∼△ABP. But △ABF is isosceles with AB=BF, hence △PBF is also isosceles with BP=BF. Thus ∠BPF=∠BFP=80∘, and since APG is collinear, ∠GPF=180∘−80∘−60∘=40∘. But ∠PGF=180∘−∠GAB−∠CBA=40∘, so △GPF is also isosceles. Therefore, DF bisects ∠GDB=60∘, which proves the claim that ∠FDB=30∘.
Next, we use this fact to prove the desired result. Note △BDF∼△AEF, because ∠AEF=180∘−70∘−80∘=30∘=∠BDF, and ∠FAB=50∘ implies ∠EAF=70∘−50∘=20∘=∠DBF. So DF/EF=BF/AF. Since ∠DFE=∠DBF+∠FDB=30∘+20∘=50∘, it follows that △ABF∼△FDE are isosceles, hence ∠AED=∠FED−∠FEA=50∘−30∘=20∘.
This is exactly the same solution as in the link I provided, but apparently was not good enough for a few people. All of it is elementary geometry.
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Buddy please give this solution to me in a simple way so that figure can be understood and a 9 OR 10 class student can understand this......as it's> too comlicated
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The proof is beautiful. Draw it out and take your time to break down every individual line - you will learn a lot from doing this!
Let the intersection of AE and BD be point X. Now use sin rule all over the place:
CE/AB =(CE/AC) (AC/AB) = (sin10/sin30) (sin80/sin20) which easily simplifies to 1. Thus CE = AB
AB/BX = sin50/sin70 = CE/BX
DB/AB = sin80/sin40
EB/AB = sin70/sin30
DC/EB = DB/EB = [(sin80)(sin30)]/ [(sin 40)(sin70)] = cos40/sin/70 = sin50/sin70 = CE/BX
Because angle DCE and angle EBX are 20, the triangles DCE and EBX are similar. Angle DEA = 20 quickly follows from some angle chase
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Its correct.. But you are not supposed to use trigonometry... Solve it by using basic triangle properties..
I had solved this problem some years ago. It had incited a lot of interest in me and was one of the main reasons of my subsequent interest in Olympiad Mathematics.
20
i don't know how but when i tried to put any number like 50,10.40.20 it fits !!!
In triangle ABE, angle AEB = 10 degrees. But angle DAE = 10 degrees also. So, the lines AC and BC must be parallel. Thus, there must be a discrepancy in the question.
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How did you come to the conclusion that ∠AEB=10∘? In △ABE, we have ∠EAB=70∘, ∠ABE=60∘+20∘=80∘, and hence ∠AEB=180∘−70∘−80∘=30∘.
its ans is 60
40????????????
Is it 40?
x=10
10
10