Given ABCD is a rhombus and DF = BE .PROVE THAT triangle ABE CONGRUENT TO TRIANGLE DFC
Attachments:
Answers
Answered by
24
<B = <D (opposite sides of parallelogram)
180 - <B = 180 - <D
or < CBE = <FDC -------------(1)
AB = DC (opposite sides of a parallelogram)
also, AB = BE (given)
so, BE = DC ----------(2)
BC = AD (opposite sides of a parallelogram)
also, AD = DF (given)
so, BC = DF -------(3)
Now, in triangle BEC and DCF,
BE = DC (from 2)
< CBE = <FDC (from 1)
BC = DF (from 3)
so, triangle BEC congruent triangle DCF (by SAS postulate)
180 - <B = 180 - <D
or < CBE = <FDC -------------(1)
AB = DC (opposite sides of a parallelogram)
also, AB = BE (given)
so, BE = DC ----------(2)
BC = AD (opposite sides of a parallelogram)
also, AD = DF (given)
so, BC = DF -------(3)
Now, in triangle BEC and DCF,
BE = DC (from 2)
< CBE = <FDC (from 1)
BC = DF (from 3)
so, triangle BEC congruent triangle DCF (by SAS postulate)
niya25:
Mark me brainlist
I will mark the option is not given
when option will be shown i will mark
thanks alot
Answered by
27
Refer to the pic
Hope this will help you out
Thanks
Attachments:
Pls mark it brainliest...I have made it short and simple
Actually leave it i done care
sorry but niya asked first
sorry
*Dont
Its fine ..Whatever u think is simple
Mark me brainlist
If you are using the SAS congruency,you have to write it in the middle between the equal sides.
Similar questions