Question

question 12 of 50
if \\(\overline{bc} \cong \overline{da}\\), then which segment is the shortest side?
image of a geometric figure with points a, b, d, c and angle measures: a=50°, angle at b in triangle abd=65°, angle at d between triangles=65°, angle at b in triangle bdc=47°, angle at d in triangle bdc=85°, angle at c=48°
options:
○ \\(\overline{dc}\\)
○ \\(\overline{bc}\\)
○ \\(\overline{ad}\\)

Explanation:

Step1: Analyze Triangle ABD

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

Step2: Analyze Triangle BCD

In $\triangle BCD$, angles are $\angle DBC = 47^\circ$, $\angle BDC = 85^\circ$, $\angle C = 48^\circ$. Sides: opposite $47^\circ$ is $DC$, opposite $48^\circ$ is $BD$, opposite $85^\circ$ is $BC$. So $DC < BD < BC$.

Step3: Use $BC \cong DA$

From Step1, $BD < AD$ (and $AD = BC$), so $BD < BC$. From Step2, $DC < BD$. Thus, $DC < BD < AD = BC$. So the shortest is $DC$? Wait, no—wait options: $\overline{DC}$, $\overline{BC}$, $\overline{AD}$. Wait, recheck. Wait in $\triangle ABD$: angles $50^\circ$, $65^\circ$, $65^\circ$: sides: $BD$ (opp $50^\circ$), $AD$ and $AB$ (opp $65^\circ$), so $BD < AD = AB$. In $\triangle BCD$: angles $47^\circ$, $85^\circ$, $48^\circ$: sides: $DC$ (opp $47^\circ$), $BD$ (opp $48^\circ$), $BC$ (opp $85^\circ$). So $DC < BD < BC$. Given $BC \cong DA$, so $BC = DA$. So $DC < BD < DA = BC$. So the shortest is $DC$? Wait the options: $\overline{DC}$, $\overline{BC}$, $\overline{AD}$. Wait but let's check again. Wait maybe I mixed up. Wait in $\triangle ABD$: $AD$ is equal to $AB$ (since angles at $B$ and $D$ are $65^\circ$). Then $BC = AD$, so $BC = AD$. Now, in $\triangle BCD$, $DC$ is opposite $47^\circ$, $BD$ opposite $48^\circ$, so $DC < BD$. In $\triangle ABD$, $BD$ is opposite $50^\circ$, $AD$ opposite $65^\circ$, so $BD < AD$. Thus, $DC < BD < AD = BC$. So the shortest is $DC$? But wait the options: first option is $\overline{DC}$. Wait but let's confirm the angles again. $\triangle ABD$: $A=50$, $ADB=65$, so $ABD=65$. So sides: $BD$ (opp 50) < $AD$ (opp 65) = $AB$ (opp 65). $\triangle BCD$: $DBC=47$, $BDC=85$, $C=48$. So sides: $DC$ (opp 47) < $BD$ (opp 48) < $BC$ (opp 85). Since $BC=AD$, so $DC < BD < AD = BC$. Therefore, the shortest is $\overline{DC}$. Wait but the options are $\overline{DC}$, $\overline{BC}$, $\overline{AD}$. So the answer should be $\overline{DC}$? Wait no, wait maybe I made a mistake. Wait the question is "which segment is the shortest side". Let's list all sides:

• $AD$: in $\triangle ABD$, angle at $A$ is $50^\circ$, angle at $D$ is $65^\circ$, so $AD$ is opposite $65^\circ$ (angle at $B$), so $AD = AB$ (since angles at $D$ and $B$ are $65^\circ$).

• $BC$: equal to $AD$ (given $BC \cong DA$).

• $DC$: in $\triangle BCD$, angle at $B$ is $47^\circ$, so $DC$ is opposite $47^\circ$, which is smaller than angle opposite $BD$ (48°) and angle opposite $BC$ (85°). So $DC < BD < BC$. And $BD < AD$ (from $\triangle ABD$: $BD$ opposite 50°, $AD$ opposite 65°). So $DC < BD < AD = BC$. Thus, $DC$ is the shortest. So the correct option is $\overline{DC}$.

Answer:

$\overline{DC}$ (the first option, $\overline{DC}$)