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In Fig. 6.15, PQR<POS = < PRT.Z PRQ, then prove thatFig. 6.15 |
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Answer» ∠PQR = ∠PRQTo prove:∠PQS = ∠PRTProof:∠PQR +∠PQS =180° (by Linear Pair axiom)∠PQS =180°– ∠PQR — (i)∠PRQ +∠PRT = 180° (by Linear Pair axiom)∠PRT = 180° – ∠PRQ∠PRQ=180°– ∠PQR — (ii)[∠PQR = ∠PRQ]From (i) and (ii)∠PQS = ∠PRT = 180°– ∠PQR∠PQS = ∠PRTHence, ∠PQS = ∠PRT |
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