# A segment BD is drawn in triangle ABC, point D lies on side AC. Segment lengths AD = 8, BD = 4, angle

A segment BD is drawn in triangle ABC, point D lies on side AC. Segment lengths AD = 8, BD = 4, angle DBC = angle DCB. Find the ratio of the area of the triangle BDC to the area ABD.

Triangle DBС is isosceles, angle DСВ = angle DBC, DС = DB = 4, АD = АС –DC = 8 – 4 = 4.

BD is the median, the median of a triangle divides it into two equal triangles.

area DBС / area ABD = 1/1.

In triangle ABC, angle C = 90 degrees, CD-height, angle A = angle a, AB = k. find Ac BC AD

by t. sinuses.

AB / SinC = BC / SinA = AC / SinB.

AB = k.

SinC = 90 ⇒ 1.

SinA = Sinα.

SinB = Sin (90-α) ⇒ Cosα.

K / 1 = BC / Sinα ⇒ BC = k * Sinα.

K / 1 = AC / Cosα ⇒ AC = k * Cosα.

CD = AC * BC / AB = ((k * Cosα) * (k * S inα)) / k = k * 1/2 * Sinα.

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