Questions: Geometry: TX > Chapter 6: Perpendicular and Angle Bisectors > Section Exercises 6.1> Exercise 18 Tell whether the information in the diagram allows you to conclude that DB=DC. Explain your reasoning. 18 No. You would need to know that AB ≅ AC. No. You would need to know that ∠BDA ≅ ∠CDA. Yes. ∠BAD ≅ ∠DAC, so DB=DC. Yes. D is on the angle bisector of ∠BAC, DB ⊥ AB and CD ⊥ AC, so DB=DC.

Geometry: TX > Chapter 6: Perpendicular and Angle Bisectors > Section Exercises 6.1> Exercise 18

Tell whether the information in the diagram allows you to conclude that DB=DC. Explain your reasoning.
18

No. You would need to know that AB ≅ AC.
No. You would need to know that ∠BDA ≅ ∠CDA.
Yes. ∠BAD ≅ ∠DAC, so DB=DC.
Yes. D is on the angle bisector of ∠BAC, DB ⊥ AB and CD ⊥ AC, so DB=DC.

Transcript text: Geometry: TX > Chapter 6: Perpendicular and Angle Bisectors > Section Exercises $6.1>$ Exercise 18 Tell whether the information in the diagram allows you to conclude that $D B=D C$. Explain your reasoning. 18 No. You would need to know that $\overline{A B} \cong \overline{A C}$. No. You would need to know that $\angle B D A \cong \angle C D A$. Yes. $\angle B A D \cong \angle D A C$, so $D B=D C$. Yes. $D$ is on the angle bisector of $\angle B A C, \overline{D B} \perp \overrightarrow{A B}$ and $\overline{C D} \perp \overrightarrow{A C}$, so $D B=D C$.

Solution

Solution Steps

Step 1: Analyze the given information

We are given that $\angle BAD \cong \angle CAD$. This means that $\overrightarrow{AD}$ is the angle bisector of $\angle BAC$. We are also given that $\overline{DB} \perp \overrightarrow{AB}$ and $\overline{DC} \perp \overrightarrow{AC}$.

Step 2: Apply the Angle Bisector Theorem

The Angle Bisector Theorem states that if a point is on the bisector of an angle, then it is equidistant from the sides of the angle. Since D lies on the angle bisector of $\angle BAC$, and $DB$ and $DC$ are the perpendicular distances from $D$ to the sides of the angle, we can conclude that $DB = DC$.

Final Answer

Yes. $D$ is on the angle bisector of $\angle BAC$, $\overline{DB} \perp \overrightarrow{AB}$, and $\overline{DC} \perp \overrightarrow{AC}$, so $DB=DC$.

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