2) Clearly BC is just extended pat of AB, and CE||DB hence BECD is a parellogram
AC congruent BD is given , and since BECD is a parellogram BD = CE
Hence, AC = CE.
Since triangle BFD is congruent to AFC, diagonals are given equal. also F will be on the point of intersection, such that BD and AC are divided in the ratio similarly.
Hence, triangle AFB is isosceles.
Can you explain part A in detail please?
Clearly BC is just extended pat of AB.
AB || CD ( since it is a trapezium )
CE || DB hence BECD is a parellogram.
oh ok thank you... & were you able to find a solution for #3?
Yes I can do it can you give a separate bid for it ????
I'm not sure what that means but ok.
I am saying can you put it in 1$ bid again....cause I have already solved one question from this sheet, which when you will best I will ge t 1$. For 2 questions I guess You are supppose to give $2
Pleas best this solution !!!
And post that questions in a separate bid I will solve it for you :-)
I usually charge $1 for one question, what i feel is correct on my part :-)
Also my complete motive is that my students understand the question well, If they do not I dont charge from them !!!
I have posted the next one Q3 !
Have answered your that question as well !!!
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