Question

In ▢ABCD, seg AD || seg BC. Diagonal AC and diagonal BD intersect each other in point P. Then show that $\frac{\mathrm{AP}}{\mathrm{PD}}=\frac{\mathrm{PC}}{\mathrm{BP}}$

Answer 1

Given: ▢ABCD is a parallelogram
To prove: $\frac{\mathrm{AP}}{\mathrm{PD}}=\frac{\mathrm{PC}}{\mathrm{BP}}$
Proof: In △APD and △CPB
∠APD = ∠CPB (Vertically opposite angles)
∠PAD = ∠PCB (Alternate angles, AD || BC and BD is a transversal line)
By AA test of similarity
△APD ∼ △CPB
$$\begin{gathered} \therefore \frac{\mathrm{AP}}{\mathrm{PC}}=\frac{{\displaystyle \mathrm{PD}}}{{\displaystyle \mathrm{PB}}}\left(\mathrm{Corresponding}\mathrm{sides}\mathrm{are}\mathrm{proportional}\right) \\ \Rightarrow \frac{\mathrm{AP}}{\mathrm{PD}}=\frac{{\displaystyle \mathrm{PC}}}{{\displaystyle \mathrm{PB}}} \end{gathered}$$
Hence proved.