Question

Q.3) ABCD is a parallelogram. P and Q are point
on the diagonal AC such that DP and Bo are
perpendiculars to AC.
Prove that DP=BO​

Answer 1

Step-by-step explanation:

consider triangle APD and triangle BQC

AD=BC ( because in parallelogram two opposites sides are equal)

angle DPA = angle BQC ( both are 90°)

angle DAP = angle BCQ ( as they are alternate angles)

therefore by ASA theorem triangle APD and triangle BQC are similar

hence DP = BO