In the accompanying diagram ab and cd are chord of the circle and intersect at e if ae=10 en=9 and ec =6 find de
To find the length of segment DE, we can use the properties of chords that intersect within a circle.
First, let's label the points more clearly. Let point A be where chord AB intersects chord CD, and point E be the intersection point of the two chords. We are given that AE = 10, EN = 9, and EC = 6.
Now, let's use the intersecting chords theorem, which states that the product of the lengths of the segments of one chord is equal to the product of the lengths of the segments of the other chord. In equation form, we have:
AE × BE = CE × DE
Substituting the given values, we have:
10 × BE = 6 × DE
To solve for DE, we need to find the value of BE. From the given information, we know that EN = 9 and NE = 9 (since NE is just AE - EN = 10 - 9 = 1).
Now, we can divide BE into BN and EN. Let x be the length of BN. Since NE = BN + EN, we have:
1 = x + 9
Solving for x, we get x = -8.
Now, let's substitute the values into the previous equation:
10 × (-8 + 9) = 6 × DE
10 × 1 = 6 × DE
10 = 6 × DE
To solve for DE, we divide both sides of the equation by 6:
DE = 10 / 6 = 5/3
Therefore, the length of segment DE is 5/3.