In the given fig., ray PQ bisects ∠ as well as ∠. State three
pairs of equal parts in ∆DPC and ∆BPC.
(only accurate answer)
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Three equal parts of
∆DPC and ∆BPC
∠DPC = ∠BPC (PQ bisects ∠DPB)
∠DCP = ∠BCP (PQ bisects ∠DCB)
PC = PC ( Common)
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