P is a point in the exterior of ∆XYZ, show that PX+PY+PZ>1/2(XY+YZ+ZX)
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Answer:
(PX + PY + PZ) > (1/2) (XY + YZ + ZX)
Step-by-step explanation:
in Δ PXY
PX + PY > XY ( sum of two sides of triangle > third side)
in Δ PXZ
PX + PZ > XZ
in Δ PYZ
PY + PZ > YZ
Adding all three
PX + PY + PX + PZ + PY + PZ > XY + XZ + YZ
=> 2(PX + PY + PZ) > XY + YZ + ZX
=> (PX + PY + PZ) > (1/2) (XY + YZ + ZX)
QED
Proved
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