In trapezoid ABCD (BC II AD), K-intersection point of the diagonals, AK: KC = 9: 4. KD-BK = 10 cm. Find BD.

univerkov | June 27, 2021 | education

Let us prove that triangles BCK and ADK are similar.

In triangles, the angle ВСК = DАК, and the angle СВК = АDК, as cross-lying angles when the secants ВD and АС intersect parallel straight lines ВС and АD. Then the triangles ВСК and АDК are similar in the first attribute, in two angles.

Then AK / СK = DK / ВK = 9/4.

Let ВK = 4 * X cm, then AK = 9 * X cm.

By condition, DK – ВK = 10 cm, then (9 * X – 4 * X) = 10.

5 * X = 10 cm.

X = 10/5 = 2 cm.

DK = 2 * 9 = 18 cm.

ВK = 2 * 4 = 8 cm.

BD = BK + DK = 8 + 18 = 26 cm.

Answer: Length BD is 26 cm.

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