19. 如图，在$△ ABC$中，$EF// AD$，$∠ 1=∠ 2$，$∠ BAC=α$. 证明：$∠ AGD=180°-α$.

19. $\because EF// AD$，$\therefore ∠ 2 = ∠ 3$.$\because ∠ 1 = ∠ 2$，$\therefore ∠ 1 = ∠ 3$.$\therefore AB// DG$.$\therefore ∠ BAC + ∠ AGD = 180°$.$\because ∠ BAC = α$，$\therefore ∠ AGD = 180 - α$

20. 如图，在四边形$ABCD$中，$AD// BC$，$∠ B=∠ D$，延长$BA$至点$E$，连接$CE$，且$CE$交$AD$于点$F$，$∠ EAD$和$∠ ECD$的平分线相交于点$P$.
(1) ① 直接写出$AB$和$CD$的位置关系：；
② 求证：$\dfrac{1}{2}∠ EAD+\dfrac{1}{2}∠ ECD=∠ APC$.
(2) 若$∠ B=70°$，$∠ E=60°$，求$∠ APC$的度数.
(3) 若$∠ APC=m°$，$∠ EFD=n°$，请你探究$m$和$n$之间的数量关系.

20. (1) ① $AB// CD$ ② 如图，过点$P$作$PQ// AB$，则$∠ EAP = ∠ APQ$，$\because AB// CD$，$\therefore PQ// CD$，$\therefore ∠ DCP = ∠ CPQ$，$\therefore ∠ EAP + ∠ DCP = ∠ APC$.$\because ∠ EAP = \frac{1}{2}∠ EAD$，$∠ DCP = \frac{1}{2}∠ ECD$，$\therefore \frac{1}{2}∠ EAD + \frac{1}{2}∠ ECD = ∠ APC$ (2) 由(1)知$AD// BC$，$AB// CD$，$\therefore ∠ EAD = ∠ B = 70°$，$∠ ECD = ∠ E = 60°$，由(1)知$∠ EAD + ∠ ECD = 2∠ APC$，$\therefore ∠ APC = \frac{1}{2}×(70°+60°) = 65°$ (3) 如图，过点$F$作$FH// AB$，则$∠ EAD = ∠ AFH$.$\because AB// CD$，$\therefore FH// CD$，$\therefore ∠ ECD = ∠ CFH$.$\therefore ∠ EAD + ∠ ECD = ∠ AFH + ∠ CFH = ∠ AFC = ∠ EFD$，由(1)知$∠ EAD + ∠ ECD = 2∠ APC$，$\therefore ∠ EFD = 2∠ APC$.$\because ∠ APC = m°$，$∠ EFD = n°$，$\therefore m = \frac{1}{2}n$