1. Since ∠CBA ≅ ∠DCB and AB ≅ BC, by the Side-Angle-Side (SAS) congruence criterion, △ABD ≅ △BCD.
2. Given that ODFE is a rectangle, all angles are right angles and opposite sides are equal. Thus, △ODF ≅ △EDF by the Side-Angle-Side (SAS) congruence criterion, as OD ≅ EF, DF is common, and ∠ODF ≅ ∠EDF.
3. Since MN ≅ MP and NQ ≅ PQ, and MQ is common to both triangles, by the Side-Side-Side (SSS) congruence criterion, △MQN ≅ △MPN.
4. Given ∠KJA ≅ ∠KJS, JK ≅ JK (common side), and KJ bisects SK, then SJ ≅ JZ (by definition of angle bisector in this context). Therefore, △JKS ≅ △JKZ by the Side-Angle-Side (SAS) congruence criterion.
5. Since ∠KND ≅ ∠LND and ND is common, and given that KNDL is a rhombus (so KN ≅ LN), by the Side-Angle-Side (SAS) congruence criterion, △KND ≅ △LND.
6. Given JK ≅ SL, ∠JKR ≅ ∠SLR, and KR ≅ LR, by the Side-Angle-Side (SAS) congruence criterion, △JKR ≅ △SLR.
7. Since KL ≅ KU and LS ≅ US, and KS is common, by the Side-Side-Side (SSS) congruence criterion, △KLS ≅ △KUS.
8. Given FG ≅ HK, FI ≅ OK, and since FGHK is a square (implied by the diagram and context), all angles are right angles. Thus, △FGI ≅ △HKO by the Side-Angle-Side (SAS) congruence criterion.
9. Since UV ≅ UW and VY ≅ WY, and UY is common, by the Side-Side-Side (SSS) congruence criterion, △UVY ≅ △UWY.
10. Given YH ≅ ZH and HK is altitude of △YZK, so ∠YHK ≅ ∠ZHK (both right angles). Since HK is common, by the Side-Angle-Side (SAS) congruence criterion, △YHK ≅ △ZHK.
Parent Tip: Review the logic above to help your child master the concept of geometry triangle proof worksheet.
