Question

the tangent to the circumcircle of triangle $wxy$ at $x$ is drawn, and the line through $w$ that is parallel to this tangent intersects $\overline{xy}$ at $z.$ if $xy = 14$ and $wx = 6,$ find $yz.$

Answer 1

Using the properties of tangents and parallel lines, the length of segment YZ is calculated to be 10.5 units by setting up a proportion based on similar triangles and solving for YZ.

Step-by-step explanation:

To solve the problem, we apply the properties of tangents and parallel lines to find the length of segment YZ in the triangle WXY with its circumcircle. Since the line through W parallel to the tangent at X intersects XY at Z, by the properties of similar triangles, WX is to WZ as XY is to YZ (alternate segment theorem).

WZ is the entire length of WX plus YZ, so we have WX + YZ = total length of WZ. We set up the proportion:

\(\frac{WX}{WZ} = \frac{XY}{YZ}\)

Given that WX = 6 and XY = 14, we can express WZ as (6 + YZ) to find the value of YZ. Solving for YZ, we find:

\(6 / (6 + YZ) = 14 / YZ\)

Cross-multiply to get:

6 * YZ = 14 * (6 + YZ)

6YZ = 84 + 14YZ

8YZ = 84

YZ = 10.5

Therefore, the length of segment YZ is 10.5 units.

Answer 2

To find the length of segment YZ, we utilize properties of similar triangles and the tangent-secant theorem. After finding the lengths of the other segments, we calculate YZ to be 8√21 / 3.

Step-by-step explanation:

We are given a triangle WXY with a circumcircle, and we draw a tangent at point X. We are told that line WZ is parallel to this tangent and hits line XY at point Z. We want to find the length of segment YZ given that XY = 14 and WX = 6.

Since WZ is parallel to the tangent at X, and WX is a radius of the circumcircle to the point of tangency, we can apply the tangent-secant theorem. This theorem states that the tangent segment squared is equal to the product of the external segment times the whole secant segment (which in this case is the sum of WX and XZ). Formally, we have WZ2 = WX * (WX + XZ).

WZ and XY are parallel, and since XY = 14, this implies XZ must be 14 - WX, or 14 - 6 which is 8. Using the tangent-secant theorem, we substitute WX = 6 into the equation to find WZ2 equals to 6 * (6 + 8), which leads us to WZ2 = 84. Therefore, WZ equals to the square root of 84, which simplifies to 2√21.

Now, because WZ is parallel to the tangent at X and intersects XY at Z, WXY and WYZ are similar triangles by the AA criterion (both have a right angle at X and share angle WXY). The ratio of the sides opposite to angle WXY in both triangles will be the same, so we can write the proportion as WX/WZ = XZ/YZ. Plugging in WX = 6, XZ = 8, and WZ = 2√21, we can solve for YZ, which yields YZ = (8 * 2√21) / 6. This simplifies to YZ = 8√21 / 3.