Challenging Geometry Puzzle: Problem 1573. Share your solution by posting it in the comment box provided.

Audience: Mathematics Education - K-12 Schools, Honors Geometry, and College Level.

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ReplyDeleteIn triangle GHF we have

Angle H=56= ∡(CGJ)-∡(CFI)

56= ½(Arc(CE)+Arc(EJ)-Arc(CI)).....(1)

In triangle KDE we have

x= angle(K)= ∡(IDE)-∡(KED)

x=½(Arc(DC)+Arc(CI)-Arc(EJ)).....(2)

Add (1) to (2) we have 56+x=½(Arc(CE)+Arc(CD))

Since AD//BE=> ∡(DAC) supplement to ∡(CBE)=> ½(Arc(CE)+Arc(CD))= 90

So x= angle(K)= 90-56= 34 degrees

Draw the common tangent CM, M on DE

ReplyDeleteNow DM = CM = ME, hence M is the centre of right triangle DCE

Let < CDE = a, < KED = b and < CDI = t = CFI

Now < KEC = b + (90 - a) = < JGC = 56 + t from which

t + a = 34 + b ....(1)

But from Triangle KDE,

< K + b = < EDI = t + a .....(2)

Comparing (1) and (2)

K = 34

Sumith Peiris

Moratuwa

Sri Lanka

Much simpler Geometry Solution

ReplyDeleteLet < DCF = u

As before < DCE = 90

< DIF = u

< DIH =- 180 - u

< EJH = 90 + u

So in quadrilateral KIHJ opposite angles J + I = (180 - u) + (90 + u) = 270

So the angles K + H = 360 - 270 = 90

Hence < K = 90 - 56 = 34

Sumith Peiris

Moratuwa

Sri Lanka