Given that ABCD is a pentagon with parallel sides AB and CD, and has four points of tangency E,F,G and H with the inscribed circle Ω at sides AB,BC,CD, and AD, respectively. If AE = 3, BE = 5, CE = 7, DE = 4, and the radius of the inscribed circle Ω is 10, what is the length of CD?
To find the length of CD, we can make use of the property that tangents drawn from an external point to a circle are equal in length.
Since AE and DE are tangent to the inscribed circle Ω at points E and H respectively, we have:
AE = DE = 3 + 4 = 7
Similarly, since BE and CE are tangent to the inscribed circle Ω at points F and G respectively, we have:
BE = CE = 5 + 7 = 12
Now, let's draw the diagonals AC and BD of the pentagon ABCD. Notice that they intersect at the center of the inscribed circle Ω.
Since AB and CD are parallel, we can consider the triangles ABE and CDE. These triangles are similar since they share the same angles and side ratio:
AE / DE = BE / CE
Substituting the given values, we have:
7 / 7 = 12 / CE
Cross-multiplying, we get:
12 * 7 = 7 * CE
CE = 12 * 7 / 7
CE = 12
Therefore, the length of CD is equal to the length of CE, which is 12.