Given: BA is congruent to DC; BE is perpendicular to AC; DF is perpendicular to AC.
Prove: BE is congruent to DF.
The image is of rhombus ABCD. There is a segment connecting A and C with points E and F each about a third of the way into the segment.
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To prove that BE is congruent to DF in the given rhombus ABCD, we can use the concept of congruent triangles. Specifically, we can use the Side-Angle-Side (SAS) congruence criterion.
Here is how we can prove that BE is congruent to DF in the given rhombus:
1. Given: BA is congruent to DC; BE is perpendicular to AC; DF is perpendicular to AC.
2. Since AC is a diagonal of the rhombus, it divides the rhombus into two congruent triangles, namely ΔABC and ΔADC.
3. Therefore, we have BA ≅ DC (given).
4. In ΔABE and ΔDFC:
- Angle BAE and angle CDF are both right angles (given).
- Angle BAE is congruent to angle CDF because they are vertical angles.
- Therefore, we have angle BAE ≅ angle CDF.
- Side BE is congruent to side DF because they are segments perpendicular to the same line segment AC.
- Therefore, we have BE ≅ DF.
5. By applying the SAS congruence criterion to ΔABE and ΔDFC, we can conclude that these triangles are congruent.
6. When two triangles are congruent, all corresponding parts (sides and angles) are congruent.
7. Therefore, BE is congruent to DF since they are corresponding sides of congruent triangles.
8. Thus, BE is congruent to DF, as desired.
By proving that BE is congruent to DF using the SAS congruence criterion, we have provided a mathematical explanation for why these line segments are congruent based on the given information.