Suggested languages for you:

Americas

Europe

Problem 100

When looking at the reflection of an object in the mirror, it seems as if the object is actually behind the mirror. The place where the object seems to be is called the virtual image. In the accompanying figure, light from the object at point \(O\) reflects off the mirror \(\underline{M N}\) at point \(B\) and strikes the eye at point \(D\). Given (by the laws of physics) that the angle of incidence equals the angle of reflection (\angleOBM $\cong \angle \mathrm{DBN}$ ) and that the total distance from the eyes to the virtual image, I, must equal the total distance traveled by the light from the eyes to the object, show that the virtual image is diametrically opposite the object with respect to the mirror.

Expert verified

In summary, we have proven that the virtual image I is diametrically opposite the object O with respect to the mirror MN using law of reflection and geometric properties. We proved that \(\triangle BOD\cong\triangle BOA\) by angle-angle-angle (AAA) congruence, which allowed us to show that the total distance from the eyes to the virtual image (DI) is equal to the total distance traveled by the light from the eyes to the object (BI). This confirms that the virtual image and the object are equidistant from the mirror and located on opposite sides, as required.

What do you think about this solution?

We value your feedback to improve our textbook solutions.

- Access over 3 million high quality textbook solutions
- Access our popular flashcard, quiz, mock-exam and notes features
- Access our smart AI features to upgrade your learning

Chapter 6

In the accompanying figure, \(D, E\), and \(F\) are the midpoints of the sides of \(\triangle \mathrm{ABC}\). If $\mathrm{AB}=\mathrm{BC}, \angle \mathrm{A} \cong \angle \mathrm{C}\(, prove that \)\mathrm{DF}=\mathrm{EF}$.

Chapter 6

Given: \(\underline{A C}\) is the perpendicular bisector of \(\underline{B D}\) at point \(O\). Prove: \(\underline{\mathrm{DC}} \cong \underline{\mathrm{BC}}\).

Chapter 6

Given: \(\angle A \cong \angle D, A E \cong D E, A C \cong \underline{D B} .\) Prove: \(\Delta A B E \cong \triangle D C E\).

Chapter 6

Given isosceles \(\Delta \mathrm{ABC}\) with $\underline{\mathrm{CA}} \cong \underline{\mathrm{CB}} . \mathrm{M}\( is the midpoint of \)\underline{A B}$ and points \(D\) and \(E\) are placed on \(\underline{C A}\) and \(\underline{C B}\), respectively, so as to make \(\underline{A D} \cong \underline{B E}\). Prove that \(\underline{M D} \cong \underline{M E}\).

Chapter 6

Given: \(\underline{A B} \cong \underline{A C}\) and $\underline{A Q} \cong \underline{A P} .\( Prove \)\angle B Q A \cong \angle C P A$.

The first learning app that truly has everything you need to ace your exams in one place.

- Flashcards & Quizzes
- AI Study Assistant
- Smart Note-Taking
- Mock-Exams
- Study Planner