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Two triangles BAC and BDC right angled at A and D respectively, are drawn on the same base BC and on the same side of BC. If AC and DB intersect at P, prove that AP × PC = DP × PB. |
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Answer» Two triangles BAC and BDC right angled at A and D respectively, are drawn on the same base BC and on the same side of BC. If AC and DB intersect at P, prove that AP × PC = DP × PB. |
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