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In adjacent figure, sides AB and AC of ∆ABC areextended to points Pand Q respectivelyAlso, <PBC< <QCB. Show that AC >AB. |
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Answer» We have ∠PBC < ∠ QCBMultiplying -1 and adding 180° both sides180° - ∠PBC > 180° - ∠ QCBSo, ∠ABC > ∠ ACB, thereforeAC > AB |
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