As shown in the figure, there is a point C on the semicircle o with the diameter of ab. through point a, make the tangent of the semicircle and intersect the extension line of BC at point D If BC = 2, root sign 3, EF = 1, find AC It has been proved that delta ADC is similar to triangle BDA The painting is not very good.

As shown in the figure, there is a point C on the semicircle o with the diameter of ab. through point a, make the tangent of the semicircle and intersect the extension line of BC at point D If BC = 2, root sign 3, EF = 1, find AC It has been proved that delta ADC is similar to triangle BDA The painting is not very good.

How can you talk up here!
Let R be the radius
According to the vertical diameter theorem in △ 0ec
R square = 3 + (R-1) square
The solution is r = 2
And in △ ABC
OE is the median line
So AC = 2oe = 4
Other conditions are superfluous