Question

question 16 of 50
if \\(\overline{ab} \cong \overline{cd}\\), then which segment is
the longest side?
triangle diagram with angles at a: 50°, at b: 65° and 47°, at d: 65° and 85°, at c: 48°
options: \\(\overline{bc}\\), \\(\overline{ad}\\), \\(\overline{dc}\\)

Explanation:

Step1: Analyze Triangle ABD

In $\triangle ABD$, angles are $50^\circ$, $65^\circ$, so the third angle $\angle ADB = 180 - 50 - 65 = 65^\circ$. So sides: $AB$ is opposite $65^\circ$, $AD$ opposite $65^\circ$, $BD$ opposite $50^\circ$. Thus $AB = AD$ (since opposite equal angles), and $AB > BD$.

Step2: Analyze Triangle BCD

In $\triangle BCD$, angles are $47^\circ$, $48^\circ$, $85^\circ$. The largest angle is $85^\circ$, so the side opposite ($BC$) is the longest in $\triangle BCD$.

Step3: Use $AB \cong CD$

We know $AB = CD$. Now compare sides: In $\triangle ABD$, $AB > BD$. In $\triangle BCD$, $BC$ is longest there. Now, let's check angles/sides. The largest angle among all related angles: in $\triangle BCD$, $\angle BDC = 85^\circ$, which is larger than angles in $\triangle ABD$ (max $65^\circ$). Wait, no—wait, the sides: $AB = CD$. Let's list all sides:

• $\triangle ABD$: $AB = AD$ (opposite $65^\circ$), $BD$ (opposite $50^\circ$)
• $\triangle BCD$: $BC$ (opposite $85^\circ$), $CD$ (opposite $47^\circ$), $BD$ (opposite $48^\circ$)

Since $AB = CD$, and in $\triangle BCD$, $BC > CD$ (because $85^\circ > 47^\circ$), so $BC > CD = AB$. And $AB = AD$, so $BC > AD$, $BC > BD$, $BC > DC$ (since $BC > CD$). Thus $BC$ is the longest.

Answer:

$\overline{BC}$ (corresponding to the first option, e.g., A. $\overline{BC}$ if options are labeled that way; based on the options, the first circle is $\overline{BC}$, so the answer is the segment $\overline{BC}$)