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Take a look at figure given below. Give it a try and when you are ready then watch the solution.
At first look this puzzle may seen very complicated but it has an easy solution. Here I will show it how...
This puzzle solution doesn't involve any type of construction. Lines given in this diagram are sufficient to get solution.
Solution:-
1) Mark AH as p and HC as q
2) Mark triangles 1, 2, 3, 4 as shown in figure
3) Mark angles for each triangles as shown in figure. (I didn't marked central angles)
Now consider Triangles (1) and (4) [AHF and BHC]:
i) Central angles are vertically opposite angles (AHF = BHC)
ii) Angles FAH = HCB (Alternative angles)
iii) Angles AFH = HBC (Alternative angles)
⇒ Triangle AHF is similar to Triangle BHC (By A-A-A Test)
⇒ p/q = 4/x ---------(1)
Similarly consider Triangles (2) and (3) [AHB and CHE]:
i) Central angles are vertically opposite angles (AHB = EHC)
ii) Angles BAH = HCE (Alternative angles)
iii) Angles ABH = HEC (Alternative angles)
⇒ Triangle AHB is similar to Triangle CHE (By A-A-A Test)
⇒ p/q = x/9 ---------(2)
From (1) and (2):
4/x = x/9
⇒ x^2 = 36
⇒ x = 6
