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Abcd is a Rhombus, Eabf is a Straight Line Such that Ea = Ab = Bf. Prove that Ed and Fc When Produced Meet at Right Angles - Mathematics

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प्रश्न

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

उत्तर

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

∴ OA = OC, OB = OD, `∠`AOD = `∠`COD = 90°

And `∠`AOB = `∠`COB = 90°

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 of 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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पाठ 13: Quadrilaterals - Exercise 13.3 [पृष्ठ ४३]

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आरडी शर्मा Mathematics [English] Class 9
पाठ 13 Quadrilaterals
Exercise 13.3 | Q 8 | पृष्ठ ४३

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