Cho tam giác $ABD$ và tam giác $IKH$ có $AB = KI,AD = KH,DB = IH.$

Nếu \(\widehat A = {60^ \circ }\), thì số đo góc $K$ là:

\({60^ \circ }\)

\({70^ \circ }\)

\({90^ \circ }\)           

\({120^ \circ }\)

\(\Delta BAD = \Delta HIK\)           

\(\Delta ABD = \Delta KHI\)

\(\Delta DAB = \Delta HIK\)           

\(\Delta ABD = \Delta KIH\)

\(\Delta BAD = \Delta HIK\)           

\(\Delta ABD = \Delta KHI\)

\(\Delta DAB = \Delta HIK\)           

\(\Delta ABD = \Delta KIH\)

\(\widehat {CAC'} > \widehat {CBC'}\)

\(\widehat {CAC'} < \widehat {CBC'}\)     

\(\widehat {CAC'} = \widehat {CBC'}\)     

\(\widehat {CAC'} = 2.\widehat {CBC'}\)

\(\widehat {BCA} = \widehat {BAC'}\)

\(\Delta ACB = \Delta BAC'\)

\(\widehat {BCA} = \widehat {ABC'}\)       

\(\Delta ACB = \Delta BC'A\)

\(\widehat {BCA} = \widehat {BAC'}\)

\(\Delta ACB = \Delta BAC'\)

\(\widehat {BCA} = \widehat {ABC'}\)       

\(\Delta ACB = \Delta BC'A\)

\(\Delta ABC = \Delta ACD\) 

\(\Delta ABC = \Delta CDA\)   

\(\Delta ABC = \Delta ADC\) 

\(\Delta ABC = \Delta CAD\)