Cumulative Assessment
Sign In
Use the Right Triangle Similarity Theorem.
See solution.
Let's start by considering the given diagram.
Let's now consider then given information regarding the diagram.
|
Right Triangle Similarity Theorem |
|
If the altitude is drawn to the hypotenuse of a right triangle, then the two triangles formed are similar to the original triangle and to each other. |
By this theorem, we can conclude that △ JPL and △ PKL are similar because PK is an altitude of △ JPL. △ JPL ~ △ PKL Let's show these triangles on the diagram, highlighting their corresponding sides, the right angles, and one pair of corresponding acute angles.
Now, since congruent triangles are similar triangles, we have that △ JPL and △ NPL are similar. △ JPL ≅ △ NPL ⇒ △ JPL ~ △ NPL Therefore, by the Transitive Property of Similarity, we can conclude that △ PKL and △ NPL are similar triangles. △ PKL ~ △ JPL △ JPL~ △ NPL ⇓ △ PKL ~ △ NPL Moreover, note that PM is an altitude of △ NPL. Therefore, by the Right Triangle Similarity Theorem, we have that △ NPL and △ NMP are also similar. △ NPL ~ △ NMP Let's show △ PKL, △ NPL, and △ NMP, highlighting corresponding sides, right angles, and corresponding acute angles.
Therefore, by the Transitive Property of Similarity, we can conclude that △ PKL and △ NMP are similar. △ PKL ~ △ NPL △ NPL~ △ NMP ⇓ △ PKL ~ △ NMP
Given: & △ JPL ≅ △ NPL & PKis an altitude of △ JPL. & PM is an altitude of △ NPL. Prove: & △ PKL ~ △ NMP We will summarize the proof we did above in a two-column table.
Statements
|
Reasons
|
1. PK is an altitude of △ JPL.
|
1. Given
|
2. △ JPL ~ △ PKL
|
2. Right Triangle Similarity Theorem
|
3. △ JPL ≅ △ NPL
|
3. Given
|
4. △ JPL ~ △ NPL
|
4. Congruent triangles are similar triangles.
|
5. △ PKL ~ △ NPL
|
5. Transitive Property of Similarity
|
6. PM is an altitude of △ NPL.
|
6. Given
|
7. △ NPL ~ △ NMP
|
7. Right Triangle Similarity Theorem
|
8. △ PKL ~ △ NMP
|
8. Transitive Property of Similarity
|