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that it lies on the bisecl12.100, AD丄CD and CBCD . If AQ = BP and DP = CQ, proveLDAQ = <CBP. |
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Answer» Given : AD⊥ CD and BC⊥ CDAQ = BP and DP = CQTo prove :∠ DAQ =∠ CBPProof :AD⊥ CD and BC⊥ CD∴∠ D =∠ C (each 90°)∵ DP = CQ (Given)Adding PQ to both sides. we getDP + PQ = PQ + CQ⇒ DQ + CPNow, in right angles ADQ and BPC∴ Hyp. AQ = Hyp. BPSide DQ = side CP∴Δ ADQ≡Δ BPC (Right angle hypotenuse side)∴∠ DAQ =∠CBP (Corresponding part of congruent triangles)Hence proved. |
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