Q.ABCD is a cyclic quadrilateral. The extended lines AB and DC intersect at point P, then— (a) PA.PB=PC.PD (b) PA.PC=PPDB. (c) PA.PC\(\lt\)PB.PD (d) PA.PB\(\gt\)PC.PD
Answer: A
In triangles PAD and PCB: \(\angle\)PAD = \(\angle\)PCB [opposite exterior angles] \(\angle\)PDA = \(\angle\)PBC [opposite exterior angles] ∴ Triangles PAD and PCB are similar. ∴ \(\cfrac{PA}{PC} = \cfrac{PD}{PB}\) ∴ \(PA \cdot PB = PC \cdot PD\)
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