Question

In the given figure, seg AC and seg BD intersect each other in point P and $\frac{\mathrm{AP}}{\mathrm{CP}}=\frac{{\displaystyle \mathrm{BP}}}{{\displaystyle \mathrm{DP}}}$. Prove that, ∆ABP ~ ∆CDP

Answer 1

Given: $\frac{\mathrm{AP}}{\mathrm{CP}}=\frac{{\displaystyle \mathrm{BP}}}{{\displaystyle \mathrm{DP}}}$
To prove: ∆ABP ~ ∆CDP
Proof: In ∆ABP and ∆DCP
$\frac{\mathrm{AP}}{\mathrm{CP}}=\frac{{\displaystyle \mathrm{BP}}}{{\displaystyle \mathrm{DP}}}$ (Given)
∠P = ∠P (Common)
By SAS test of similarity
$\frac{\mathrm{AP}}{\mathrm{CP}}=\frac{{\displaystyle \mathrm{BP}}}{{\displaystyle \mathrm{DP}}}$