Chapter 6.6, Problem 32PPS

To determine

To prove:

The following statement with paragraph proof:

Theorem 6.26: $\angle \mathrm{B}\cong \angle \mathrm{D},\angle \mathrm{B}\mathrm{A}\mathrm{D}\cancel{\cong }\angle \mathrm{B}\mathrm{C}\mathrm{D}$

Explanation of Solution

Given information:

  $\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}$ is kite

  $\overline{\mathrm{A}\mathrm{C}}\cong \overline{\mathrm{B}\mathrm{D}}$

Proof:

Here, as per definition of kite $\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}$

  $\overline{\mathrm{A}\mathrm{B}}\cong \overline{\mathrm{A}\mathrm{D}}$ and $\overline{\mathrm{B}\mathrm{C}}\cong \overline{\mathrm{C}\mathrm{D}}$ . According to the reflexive property of congruence, $\overline{\mathrm{A}\mathrm{C}}\cong \overline{\mathrm{A}\mathrm{C}}$ . By side-side-side theorem $\mathrm{\Delta }\mathrm{A}\mathrm{B}\mathrm{C}\cong \mathrm{\Delta }\mathrm{A}\mathrm{D}\mathrm{C}$ . Know that the corresponding parts of congruent triangles are congruent so $\angle \mathrm{C}\cong \angle \mathrm{D}$

  

  $\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}$ is parallelogram is it has angle congruence $\angle \mathrm{B}\mathrm{A}\mathrm{D}\cong \angle \mathrm{B}\mathrm{C}\mathrm{D}$ . It is not true because $\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}$ is kite as per the given information. Thus, $\angle \mathrm{B}\cong \angle \mathrm{D},\angle \mathrm{B}\mathrm{A}\mathrm{D}\cancel{\cong }\angle \mathrm{B}\mathrm{C}\mathrm{D}$ for $\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}$ as $\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}$ is kite not a parallelogram.

Hence proved