Through the vertex a of △ ABC, make AE ⊥ AB, ad ⊥ AC, make AE = AB, ad = AC, BD and CE intersect at point O. prove: (1) EC = BD (2) EC ⊥ BD

Through the vertex a of △ ABC, make AE ⊥ AB, ad ⊥ AC, make AE = AB, ad = AC, BD and CE intersect at point O. prove: (1) EC = BD (2) EC ⊥ BD

It is proved that the intersection of AB and CE is set as G ∵ AE ⊥ AB, ad ⊥ AC ≁ BAE ≁ CAD = 90 ≁ AEC + ≌ age = 90 ≂ bad = ≂ BAC + ≁ CAD, ≁ CAE = ≂ BAC + ≂ BAE ≂ bad = ≌ CAE ≂ AE = AB, ad = AC ≌ bad ≌ EAC (SAS) ≌ EC = BD, ≌ abd = ≌ AEC ≂ age = BGO

Ad is the middle line of the triangle ABC, CE is parallel to AB, ad bisects the angle BAE

It is proved that BF = EC on Ba is connected with be and FC. Since BF is parallel and equal to CE, CeBF is a parallelogram,
D is the midpoint of diagonal BC, and the other diagonal EF must intersect BC at D, that is, D is the midpoint of EF, so Ed = FD;
It is also known that ad = ad; ead = fad
sin∠AED=(ADsin∠EAD)/DE=(ADsin∠FAD)/DF=sin∠AFD,
∴∠AED=∠AFD.
∴AF=AE,∴AB=AF+BF=AE+EC.
Therefore, it is difficult to prove