Ap Dp

dP Gold ap dp In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ Show that: i) ΔAPD ≅ ΔCQB ii) AP = CQ iii) ΔAQB ≅ ΔCPD iv) AQ = CP apkpure open AP and DP brackets are set ranges that give bonus AP or extra damage reduction For example, once you hit 100 AP, you get an extra base 5 AP

aposta Given: Seg AC and seg BD intersect each other in point P and AP CP BP DP AP CP = BP DP To prove: ∆ABP ~ ∆CDP Proof: In ∆ABP and ∆CDP,