# ABCD is a rhombus with P,Q,R as mid points of AB,BC,CD respectively. Prove that PQ is perpendicular to QR

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by mayurKonda

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by mayurKonda

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and it inersects PQ at F and QR at G

QR//BD,hence,QG//EF

llly, FQ//EG

i.e.EFGQ is a llgm

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

hence,<FEG=90

thus,<FQG=90(opposite angles of a llgm are equal)

so,PQR=90

i.e.PQ is perpendicular to QR