A geometry problem by Ossama Ismail

Geometry Level 2

Given a circle with diameter $\overline {AC}=8,$ line segment $\overline {FB} = 4.5$ is tangent to the circle at point $B,$ and $\overline {FD}$ is perpendicular to $\overline {AC}.$

Find the sum of lengths $\overline {FE} + \overline {FB}.$

8 9.5 9 8.5

1 solution

Roger Erisman
Mar 9, 2017

Let center of circle be O.

Then OA = OB = radius of circle, so angle OAB = angle OBA. Call this angle y.

Call angle AED = x, so x + y = 90

Angle FEB = x because it is vertical to AED, and angle OBF = 90 since a radius to the tangent line is perpendicular.

Therefore angle FBE must equal x since FBO = 90 and OBE = y and we know that x + y = 90.

Since FEB = x and FBE = x, the triangle is isosceles and FE = FB so FE + FB = 4.5 + 4.5 = 9

Quad. $CBED$ is a cyclic quadrilateral. => Angle FEB = Angle BCA = Angle FBE. .'. FE = FB = 4.5 + 4.5 = 90

Vishwash Kumar ΓΞΩ - 4 years, 3 months ago

A geometry problem by Ossama Ismail

1 solution

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