Math Olympiad | A Very Nice Geometry Problem

[ludmilaivanova]

@ 8:33 since EF is parallele to AC and goes through the middle of BC, it is a middle line of a triangle and it is equal to the half of the base(AC).

[toninhorosa]

Dear teacher. I like the way you find to solve this problem. Congratulation!!!
👏👏👏👏👏👏

[michaeldoerr]

The answer is 30°. This is similar to to now 5 or more of the problems on this channel. I am finally gotten the HL similarity right finally and looks like I will have to use this for more practice. I will have to watch it again so that I can make this one of the geometry that I can show impeccable comprehension of!!!

[harikatragadda]

4:04 Construct ∆AOE Congruent to ∆ADE under AE and join OC.
∆OCE is Congruent to ∆AOE and ∆AOC is Isosceles.
Since ∠OAC =∠OCA =60°, ∆AOC is in fact an Equilateral Triangle.
Thus, AC = AD and ∆CAD is an Isosceles Right Triangle,
with ∠ACD = 45°.
Hence, θ= 75-45=30°

[solimana-soli]

Very nice! I found a much longer solution with trig.

[RealQinnMalloryu]

{15°B+15°D+90E}=120°BDE {120°BDE+60°}=180°BDE 2^90 2^3^30 3 2^3^3^10 2^1^3^2^5 1^1^3^2^1 3^2 (BDE ➖ 3BDE+2).

[giuseppemalaguti]

θ=30...uso sempre la stessa formula del seno