Cho hai tam giác $ABC$ và $DEF$ có $AB = EF;\,BC = FD;AC = ED;$ $\widehat A = \widehat E;\widehat B = \widehat F;\widehat D = \widehat C$. Khi đó

$\widehat A = \widehat D = {105^0};\,\widehat C\, = \widehat F = {40^0}.$

$\widehat A = \widehat D = {90^0};\,\widehat C\, = \widehat F = {50^0}.$

$\widehat A = \widehat D = {95^0};\,\widehat C\, = \widehat F = {40^0}.$

$\widehat A = \widehat D = {40^0};\,\widehat C\, = \widehat F = {95^0}.$

$4\,cm$

$6\,cm$

$8\,cm$

$10\,cm$

$\widehat A < \widehat C < \widehat B$

$\widehat A > \widehat B = \widehat C$

$\widehat A > \widehat B > \widehat C$

$\widehat A > \widehat C > \widehat B$

$PQ = DE$

$PR = EF$

$\widehat Q = \widehat E$

$\widehat D = \widehat P$

$\widehat C = {45^0}$

$\widehat B = {45^0}$

$\widehat A = {45^0}$

$\widehat C = {90^0}$

$\Delta MNP = \Delta IJK$

$\Delta MNP = \Delta IKJ$

$\Delta MNP = \Delta KIJ$

$\Delta MNP = \Delta JKI$

$\widehat D = {80^0};\,\widehat K = {40^0}$

$\widehat D = {40^0};\,\widehat K = {60^0}$

$\widehat D = {60^0};\,\widehat K = {80^0}$

$\widehat D = {40^0};\,\widehat K = {80^0}$