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.
4 years, 6 months ago
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 \( \angle FAB = 50^\circ \). Then we first show that \( \angle FDB = 30^\circ \). 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 \( \triangle DGP \sim \triangle ABP \). But \( \triangle ABF \) is isosceles with \( AB = BF \), hence \( \triangle PBF \) is also isosceles with \( BP = BF \). Thus \( \angle BPF = \angle BFP = 80^\circ \), and since \( APG \) is collinear, \( \angle GPF = 180^\circ - 80^\circ - 60^\circ = 40^\circ \). But \( \angle PGF = 180^\circ - \angle GAB - \angle CBA = 40^\circ \), so \( \triangle GPF \) is also isosceles. Therefore, \( DF \) bisects \( \angle GDB = 60^\circ \), which proves the claim that \( \angle FDB = 30^\circ \).
Next, we use this fact to prove the desired result. Note \( \triangle BDF \sim \triangle AEF \), because \( \angle AEF = 180^\circ - 70^\circ - 80^\circ = 30^\circ = \angle BDF \), and \( \angle FAB = 50^\circ \) implies \( \angle EAF = 70^\circ - 50^\circ = 20^\circ = \angle DBF \). So \( DF/EF = BF/AF \). Since \( \angle DFE = \angle DBF + \angle FDB = 30^\circ + 20^\circ = 50^\circ \), it follows that \( \triangle ABF \sim \triangle FDE \) are isosceles, hence \( \angle AED = \angle FED - \angle FEA = 50^\circ - 30^\circ = 20^\circ \).
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.
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
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
Its correct.. But you are not supposed to use trigonometry... Solve it by using basic triangle properties..
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.
How did you come to the conclusion that \( \angle AEB = 10^\circ \)? In \( \triangle ABE \), we have \( \angle EAB = 70^\circ \), \( \angle ABE = 60^\circ + 20^\circ = 80^\circ \), and hence \( \angle AEB = 180^\circ - 70^\circ - 80^\circ = 30^\circ \).
i don't know how but when i tried to put any number like 50,10.40.20 it fits !!!
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.
its ans is 60
Apr 06, 2013
Is it 40?