In the triangle ABC, the sides AB 2 AC 4 are known in what ratio the circle passing

In the triangle ABC, the sides AB 2 AC 4 are known in what ratio the circle passing through the vertices BC and the middle AB divides the side AC.

Sides AB and AC of triangle ABC are secant lines drawn from one point.

Point M is the middle of AM, then AM = AB / 2 = 2/2 = 1 cm.

Let the length of the segment AK = X cm.

By the property of secants drawn from one point, the product of the length of one secant to its outer part is equal to the product of the second secant to its outer part.

AB * AM = AC * AK.

2 * 1 = 4 * H.

X = 2/4 = 1/2.

Then CK = AC – AK = 4 – 0.5 = 3.5 cm.

Then AK / CK = 0.5 / 3.5 = 1/7.

Answer: The circle divides the AC side in a ratio of 1/7.