Line BAP = BDF; So BP = BD+DF =BD+DX; (1) Line DBT=DCH DB+BT=DC+CH ; So DB+BY=DC+CH ; So DB= DC+CH-BY (2) From (1), (2); BP = BF= DX + DC+CH-BY =DX+DC-(BC-YC)+CH=XC-BC + 2CH= AX –AX+XC-BC+2CH =AC-AX- BC+2CH= AC-AX- BA+2CH =AC+2CH-BP So BP=1/2 AC+CH= BU BP=BF=BU; Because TGxBI = BUxGI; So GHXBI =BFxGI; So BI/BF = GI/GH So BG//FH
https://photos.app.goo.gl/51ioXWMypPWvkQf49
ReplyDeleteExtend HG and FDB, intersect at I;
Draw BU, GT as in the photo.
Line BAP = BDF;
So BP = BD+DF =BD+DX; (1)
Line DBT=DCH
DB+BT=DC+CH ;
So DB+BY=DC+CH ;
So DB= DC+CH-BY (2)
From (1), (2);
BP = BF= DX + DC+CH-BY =DX+DC-(BC-YC)+CH=XC-BC + 2CH= AX –AX+XC-BC+2CH
=AC-AX- BC+2CH= AC-AX- BA+2CH =AC+2CH-BP
So BP=1/2 AC+CH= BU
BP=BF=BU;
Because TGxBI = BUxGI;
So GHXBI =BFxGI;
So BI/BF = GI/GH
So BG//FH