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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.

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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.

प्रमेय
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उत्तर


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

∴ OA = OC

OB = OD

∠AOD = ∠COD = 90°

And ∠AOB = ∠COB = 90°

In ΔBDE, A and O are midpoints of BE and BD, respectively

OA || DE

OC || DG

In ΔCFA, B and O are midpoints 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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अध्याय 13: Quadrilaterals - Exercise 13.3 [पृष्ठ ४३]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 9
अध्याय 13 Quadrilaterals
Exercise 13.3 | Q 8 | पृष्ठ ४३
नूतन Mathematics [English] Class 9 ICSE
अध्याय 9 Mid-point Theorem
Exercise 9A | Q 16. | पृष्ठ १९६

वीडियो ट्यूटोरियलVIEW ALL [1]

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