Two equal pillars AB and CD are standing on the either side of the road as shown in the figure.
If AF = CE, then prove that BE = FD.
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Answered by
66
hey dear your answer is here..
here
AB = CD
AF = CE
angle B = angle D
so bye SSA both triangle ABF and CDE are congruent.
by cpct
BF = ED
BF - EF = ED - EF
BE=FD
here
AB = CD
AF = CE
angle B = angle D
so bye SSA both triangle ABF and CDE are congruent.
by cpct
BF = ED
BF - EF = ED - EF
BE=FD
navi8526:
chal hat
Answered by
22
From the figure, you will find out two similar triangles i.e. △ABF = △CDE.
In these two triangles, you will also discover ∠B =∠D = 90°, side AF – side CE, and side AB = side CD.
According to the RHS rule, △ABF = △CDE are concurrent. As per CPCT (Corresponding parts of congruent triangles), side BF = side DE.
When you subtract common length EF from both sides, we obtain => BF – EF = DE – EF => BE – FD.
Hence, it is proved.
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