Answer:
See explanation
Step-by-step explanation:
PQRS is a parallelogram, then
[tex]PQ\parallel RS\\ \\QR\parallel SP[/tex]
and
[tex]PQ\cong RS\\ \\QR\cong SP[/tex]
Also
[tex]\angle P\cong \angle R\\ \\\angle Q\cong \angle S[/tex]
1. Consider triangles PQA and RQA, where A is the diagonals intersection point. In these triangles,
[tex]QA\cong QA\ [\text{Reflexive property}][/tex]
[tex]\angle PQA\cong \angle RQA\ [\text{Because QS is angle bisector}][/tex]
[tex]\angle QPA\cong \angle QRA\ [\text{Diagonal PR divides congruent angles in congruent halves}][/tex]
Hence, by AAS postulate triangles PQA and RQA are congruent. Congruent triangles have congruent corresponding parts, so
[tex]PQ\cong RQ[/tex]
2. Consider triangles PSA and RSA. In these triangles
[tex]SA\cong SA\ [\text{Reflexive property}][/tex]
[tex]\angle PSA\cong \angle RSA\ [\text{Because QS is angle bisector}][/tex]
[tex]\angle SPA\cong \angle SRA\ [\text{Diagonal PR divides congruent angles in congruent halves}][/tex]
Hence, by AAS postulate triangles PSA and RSA are congruent. Congruent triangles have congruent corresponding parts, so
[tex]PS\cong RS[/tex]
3. Since [tex]PQ\cong RS[/tex] and [tex]PQ\cong RQ[/tex], you have [tex]RS\cong RQ[/tex]
Therefore,
[tex]PQ\cong QR\cong RS\cong SP[/tex]
Parallelogram with all congruent sides ia a rhombus.
