Q.In cyclic quadrilateral ABCD, the extended sides AB and DC intersect at point P. Given: PA = 6 cm, PB = 2 cm, and PD = 8 cm Find: The length of PC. (a) 3 cm (b) 1.5 cm (c) 4.5 cm (d) 6 cm
Answer: B
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. ∴ \(\cfrac{PA}{PC} = \cfrac{PD}{PB}\) i.e., \(\cfrac{6}{PC} = \cfrac{8}{2}\) ⇒ \(PC = 1.5\) ∴ PC = 1.5 cm.
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