Question

10. ABCD is a rhombus EABF is a straight line such that EA = AB = BF. Prove that ED and FC
when produced meet a right angles.
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Answer 1

Answer:

We know that the diagonals of a rhombus are perpendicular bisector of each other.

∴OA=OC,OB=OD,∠AOD=∠COD=90  

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and, ∠AOB=∠COB=90  

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In ΔBDE, A and O are mid - points of BE and  

BD respectively.

∴OA∥DE

⇒OC∥DG

In ΔCFA,B and O are mid - points AF and AC respectively  

∴OB∥CF

⇒OD∥GC

Thus, in quadrilateral DOCG, we have  

OC∥DG and OD∥GC  

⇒DOCG is a parallelogram.

∴∠DGC=∠DOC

⇒∠DGC=90  

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hope it is useful

Answer 2

Answer:

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