In the ABCD trapezoid, the lateral side AB is perpendicular to the base of the BC. The circle passes through points C

| February 6, 2021 | education

In the ABCD trapezoid, the lateral side AB is perpendicular to the base of the BC. The circle passes through points C and D and touches line AB at point E. Find the distance from point E to straight line CD, if AD = 4, BC = 3.

Let’s construct the perpendicular EH to the СD, which is our required distance, and the perpendicular СM to the AD. DM = AD – BC = 4 – 3 = 1 cm.

Let’s extend the lateral sides AB and СD until they intersect at point K.

Let the length of the СD = X cm.

Triangles KBC and СDM are similar in acute angle.

KС / ВС = СD / DM.

KС = BC *СD / DM = 3 * X cm.

KE – tangent, KD secant, then TE^2 = KD * KС = 4 * X * 3 * X = 12 * X^2.

TE = 2 * √3 * X cm.

Triangles KEН and KBC are similar in acute angle, then:

EH / KE = BC / KС.

ЕН = КЕ * ВС / КС = 2 * √3 * X * 3/3 * X = 2 * √3 cm.

Answer: From point E to straight СD 2 * √3 cm.



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