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Standard X

Mathematics

AAA Similarity

In given figu...

Question

In given figure 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 $A P \times P C = D P \times P B$

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Solution

In $△ A P B$ and $△ D P C$ , we have

$∠ A = ∠ D = 90^{\circ}$ and

$∠ A P B = ∠ D P C$ [Vertically opposite angles]

Thus, by AA-criterion of similarity, we have

$△ A P B \sim △ D P C$

$\Rightarrow$ $\frac{A P}{D P} = \frac{P B}{P C}$

$\Rightarrow$ $A P \times P C = D P \times P B$ $[H e n c e p r o v e d]$

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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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In given figure 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

In △APB and △DPC, we have ∠A=∠D=90∘ and ∠APB=∠DPC [Vertically opposite angles]Thus, by AA-criterion of similarity, we have △APB∼△DPC⇒ APDP=PBPC⇒ AP×PC=DP×PB [Hence proved]

In given figure 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 $A P \times P C = D P \times P B$