Question

In □ABCD seg AD||Seg BC, diagonal AC and BD intersect each other in point P then show that AP÷PD = PC÷BP​

Answer 1

Step-by-step explanation:

prove the two triangles similar APD and BPC

prove by stating

angle p = angle p (common)

angle ADP= angle PBC

By Angle Angle theorem (AA) the two triangles are similar now so

AP/PD = PC/BP