Bunuel wrote:

In figure below, equilateral triangle ABC is inscribed in circle O, whose radius is 4. Altitude BD is extended until it intersects the circle at E. What is the length of DE?

A. 1

B. \(\sqrt{3}\)

C. 2

D. \(2\sqrt{3}\)

E. \(4\sqrt{3}\)

Attachment:

eIdABed 2018.07.09.jpg [ 33.01 KiB | Viewed 535 times ]
I. \(r = \frac{2}{3} h\)BD is an altitude (= median -- equilateral triangle)

The medians of an equilateral triangle intersect (= centroid) at the center of the circumscribed circle

The centroid divides the median into a ratio of \(2x:1x\)

Median = Altitude, \(h = 3x\)

The radius of the circle is \(\frac{2}{3}\) the altitude of the triangle

\(r = \frac{2}{3}h\)

\(h =\frac{3}{2}r\)

BE is the diameter of the circle = \(2r = 8\)

Height, BD, =\(\frac{3}{2} r =(\frac{3}{2}*4)= 6\)

(BE - BD) = DE

\((8 - 6) = 2\)

\(DE = 2\)Answer C

II. 30-60-90 trianglesThe medians (= altitudes) of an equilateral triangle divide it into six congruent 30-60-90 triangles.

∆ OAD is one of those triangles

-- BD is an altitude

-- equilateral triangle: altitude = median = perpendicular bisector of vertex and base

The ratio of sides opposite 30-60-90 angles is \(x : x\sqrt{3}: 2x\)

OD corresponds with \(x\)

AD corresponds with \(x\sqrt{3}\)

OA corresponds with \(2x\)

OA is a radius. OA = \(4\)

OA, opposite the 90° angle = \(2x\)

OA is twice the length (\(2x:x\)) of OD.

OD = \(2\)

OE = \(4\) (radius)

The length of DE is just (OE - OD) = \((4 - 2) = 2\)

\(DE = 2\)Answer C

_________________

In the depths of winter, I finally learned

that within me there lay an invincible summer.

-- Albert Camus, "Return to Tipasa"