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Question Number 207381 by efronzo1 last updated on 13/May/24

Answered by mr W last updated on 13/May/24

((BD∙CE)/(AC∙DE)) =(((BE+DE)∙CE)/((AE+CE)∙DE)) =(((((BE)/(DE))+1)∙CE)/((((AE)/(CE))+1)∙DE)) =((((BC)/(AD))+1)/(((BC)/(AD))+1)) =((x+1)/(x+1))=1

BD⋅CEAC⋅DE=(BE+DE)⋅CE(AE+CE)⋅DE=(BEDE+1)⋅CE(AECE+1)⋅DE=BCAD+1BCAD+1=x+1x+1=1

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