Challenging Geometry Problem 1604. 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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It is the square root of85.
ReplyDeleteAlhamdolillah.
ReplyDeleteFirst, we use triangle similarity, angle between tangent line and chord, chasing angles. After that, we can get the key trick which is showing that DG is tangent with the small circle. After that, by using tangent secant thereom, we can form two equations and find GD which is the square root of 85.
How DG is tangent
DeleteAlhamdulillahirst,
ReplyDeleteFirst we use triangle similarity, angle between tangent line and chord, chasing angles... After that, we can get the key trick which is showing that DG is tangent with the small circle. After that, by using tangent secant thereom, we can form two equations and find GD which is equal to square root of 85.
just finish the question by power of point bashing
ReplyDeleteIt's easy to see that CG//FG.
ReplyDeleteIf the intersection of the extension of GF and the extension of DF is G',
the quadrilateral CGDG' is a parallelogram. (The midpoint of the line segment GG' is A.)
From this and the tangent theorem, we can state that ∠FCA = ∠CGA = ∠AG'D,
and we can state that the four points ACG'F are points on the same circumference.
Therefore, ∠GDA = ∠ACG' = ∠AFD
From this, GD is a tangent to the second circle.
FB×GF = 49
From this and the fact that FB×FG = 36,
FG = √85
Excellent Solution
DeleteWell done