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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TopNewestA 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 – Hero P. · 4 years, 1 month ago

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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. – Hero P. · 4 years, 1 month ago

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– Anmol Mishra · 4 years, 1 month ago

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 comlicatedLog in to reply

– Gabriel Wong · 4 years, 1 month ago

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!Log in to reply

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 – Gabriel Wong · 4 years, 1 month ago

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– Vasavi GS · 4 years, 1 month ago

Its correct.. But you are not supposed to use trigonometry... Solve it by using basic triangle properties..Log in to reply

20 – Ngô Ngọc Mai Thy · 4 years, 1 month ago

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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. – Aasif Khan · 4 years, 1 month ago

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– Hero P. · 4 years, 1 month ago

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 \( \angle AEB = 180^\circ - 70^\circ - 80^\circ = 30^\circ \).Log in to reply

i don't know how but when i tried to put any number like 50,10.40.20 it fits !!! – Omar Ali · 4 years, 1 month ago

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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. – Sambit Senapati · 4 years, 1 month ago

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its ans is 60 – Varun Narayan · 4 years, 1 month ago

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– Giovanni Mounir · 4 years, 1 month ago

180-70-60=70??????????????????????????Log in to reply

Is it 40? – Jordi Bosch · 4 years, 1 month ago

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40???????????? – Amit Dogra · 4 years, 1 month ago

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10 – Aman Jain · 4 years, 1 month ago

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x=10 – Giovanni Mounir · 4 years, 1 month ago

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10 – Priyansh Daushalya · 4 years, 1 month ago

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