Question

question 3 (multiple choice worth 1 points) (02.04r mc) triangles abd and cbd are shown. if m∠abd = 100°, what is the relationship between ad and cd? ad + dc < ac cd = ad cd > ad cd < ad

Explanation:

Step1: Observe the figure

We see that AB = BC (marked by the equal - length symbols).

Step2: Consider the angles

Given \(m\angle ABD=100^{\circ}\), then \(m\angle DBC = 180 - 100=80^{\circ}\) since \(\angle ABD\) and \(\angle DBC\) are a linear - pair.

Step3: Apply the law of cosines

In \(\triangle ABD\), \(AD^{2}=AB^{2}+BD^{2}-2\cdot AB\cdot BD\cdot\cos\angle ABD\). In \(\triangle CBD\), \(CD^{2}=BC^{2}+BD^{2}-2\cdot BC\cdot BD\cdot\cos\angle DBC\). Since \(AB = BC\), and \(\cos\angle ABD=\cos100^{\circ}<0\), \(\cos\angle DBC=\cos80^{\circ}>0\).
Let \(AB = BC = x\) and \(BD = y\). Then \(AD^{2}=x^{2}+y^{2}-2xy\cos100^{\circ}=x^{2}+y^{2}+ 2xy|\cos100^{\circ}|\) and \(CD^{2}=x^{2}+y^{2}-2xy\cos80^{\circ}\).
Since \(|\cos100^{\circ}|=\cos80^{\circ}\), we have \(AD^{2}>CD^{2}\) (because of the positive term \(2xy|\cos100^{\circ}|\) in the formula for \(AD^{2}\)). Taking the square - root of both sides (and considering non - negative lengths), we get \(AD>CD\).

Answer:

CD < AD