如图，已知AD⊥BC于D，FG⊥BC垂足分别为D，G，且∠1=∠2，∠C=50°，求∠EDC的度数．

证明：∵AD⊥BC，FG⊥BC，

∴∠ADC=∠FGC=90°（                  ▲                  ）．

∴                  ▲                  //FG（                  ▲                  ）．

∴∠1=∠3，

又∵∠1=∠2，

∴∠2=∠3（                  ▲                  ）．

∴DE//                  ▲                   ． （                  ▲                  ）．

∴∠EDC+∠C=180°（                  ▲                  ）．

∵∠C=50°．

∴∠EDC=                  ▲                   ．