In the figure, seg AC and seg BD intersect each other in point P and AP/CP = BP/DP. Prove that, ΔABP ~ ΔCDP.

Answer: \(\triangle ABP \sim \triangle CDP\) by the SAS test of similarity.

In \(\triangle ABP\) and \(\triangle CDP\):

• \(\frac{AP}{CP} = \frac{BP}{DP}\) (given)
• \(\angle APB \cong \angle CPD\) (vertically opposite angles)

Two sides of one triangle are proportional to two sides of the other and the included angles are congruent.

\(\therefore \triangle ABP \sim \triangle CDP\) (SAS test of similarity). Hence proved.