Can you solve this 10th grade tricky geometry problem 

Great!! Very nice explanation! An alternative solution is applying II Euclid's theorem on ADB and ACB right triangles considering their heights DH and CK. Triangle DEH is an 30°,60°,90° right triangle so setting EH = a, follows that DE = 2a and DH = a√ 3, then (a√ 3)² = (4 - a)*(12 + a) (Euclid's th. on ADB) a = √ 13 - 1 => 2a = DE = 2√ 13 - 2 Triangle CEK also is an 30°,60°,90° right triangle so setting EK = b, follows that CE = 2b and CK = b√ 3, then (b√ 3)² = (12 - b)*(4 + b) (Euclid's th. on ABC) b = √ 13 + 1 => 2b = CE = 2√ 13 + 2 Now applying cosine law on DEC with angle DEC = 60° then cos 60° = 1/2 we can write: X² = (2√ 13 + 2)² + ( 2√ 13 - 2)² - 2*(2√ 13 + 2)*(2√ 13 - 2)*1/2 X² = 64 => X = 8

Il raggio R=8,EC=6 col teorema dei seni..l'equazione risolutiva è quella degli angoli in D e C come termini del raggio...essi sono ovviamente uguali .arcsin 3√3/x-arcsin√3/4=120-arcsin3√3/x+arcsin√3/4….risulta facilmente x=8

This is one of the semi-circle theorums I was taught many years ago. Providing the two angles either side of the angle DEC are the same (i.e. the bisector of DEC is perpendicular to AB) then X remains the same lenght regardless of where E is along the line AB, so it equals the length you'd get if E was at the circle's origin. In this instance DEC is 60° so X equals the radius. If the outer angles are 45° and DEC 90° then X would be radius * ✓2. Simple.

When we find inscribed angleDFC=30, we can use 2R= x/Sin 30. R=8, so x=8

Can be solved using EPT (Extension of Pythagoras Theorem) Let M be the center. Hence, MD = MC = 8. In ∆DEM, DM = 8 = side opposite to 60°. By EPT, DM^2 = EM^2 + ED^2 - EM*ED Hence, 64 = 16 + DE^2 - 4*DE Solving this quadratic we get, DE = √52 + 2 Similarly, in ∆CEM, CM is side opposite vto 120°. Hence, by EPT, CM^2 = EC^2 + EM2 + EC*EM Hence, 64 = EC^2 + 16 + 4*CE Solving this quadratic gives CE = √52 - 2 In ∆CDE, CD = x is opposite to 60°. Hence by EPT, x^2 = CE^2 + DE^2 - CE*DE Substituting above values of CE and DE, We get x = 8

I used cosine law 7 times (two times it was the special case of cosine law - Pythagorean identity) ACB right triangle and Pythagorean identity in this triangle Cosine law in triangle AEC and in triangle CEB with supplementary angles (120 and 60) Add up results from cosine law and i have quadratic equation ADB right triangle and Pythagorean identity in this triangle Cosine law in triangle AED and in triangle DEB with supplementary angles (60 and 120) Add up results from cosine law and i have quadratic equation Cosine rule in triangle EDC

asnwer= 9.5 cm 120

8

WTF? Diameter is 4 + 12 = 16, radius is 8, so EO is 8 - 4 = 4. The problem is that EO is also supposed to equal 8 / tan (60°), or 8 × tan (30°), which is 4,6188. Unless I'm missing something, this is the worst blunder I've ever seen in a math video.

DC=2\/15!

I think this is a simple coordinate geometry problem.