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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