The problem involves finding the length of the third side of a triangle using the Angle Bisector Theorem.
Given:
The Angle Bisector Theorem states that an angle bisector of a triangle divides the opposite side into two segments that are proportional to the other two sides of the triangle.
For $\Delta\text{XYZ}$ and angle bisector XW:
$ \frac{\text{YW}}{\text{WZ}} = \frac{\text{XY}}{\text{XZ}} $We are given the ratio $\frac{\text{YW}}{\text{WZ}} = \frac{4}{5}$ and the side length $\text{XY} = 12\text{ cm}$.
Substitute these values into the Angle Bisector Theorem equation:
$ \frac{4}{5} = \frac{12\text{ cm}}{\text{XZ}} $To find $\text{XZ}$, rearrange the equation:
$ \text{XZ} = \frac{12\text{ cm} \times 5}{4} $Calculate the result:
$ \text{XZ} = \frac{60\text{ cm}}{4} $ $ \text{XZ} = 15\text{ cm} $The length of the third side, $\text{XZ}$, is $15\text{ cm}$.
What is the foot of the altitude from the vertex A of the triangle ABC?
In ΔABC, D is a point on BC such that ∠ADB = 2∠DAC, ∠BAC = 70° and ∠B = 56°. What is the measure of ∠ADC?
In ΔABC, ∠A = 66° and ∠B = 50 °. If the bisectors of ∠B and ∠C meet at P, then ∠BPC – ∠PCA = ?
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