Question

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

Answer 1

Answer:

ΔABP~ΔCDP

Step-by-step explanation:

as line AC & BC intersect each other at P

=> ∠APB = ∠CPD

AP/CP = BP/DP = K

=> AP =  K * CP

& BP = K * DP

inΔ ABP

AB² = AP² + BP² - 2AP.BPCos∠APB

putting AP = K * CP & BP = K * DP

=> AB² = K²CP² + KDP² - 2K*CP.*K*DPCos∠APB

=> AB² = K² (CP² + DP² - 2CP*DPCos∠APB)

∠APB = ∠CPD

=>  AB² = K² (CP² + DP² - 2CP*DPCos∠CPD)

in Δ CDP CP² + DP² - 2CP*DPCos∠CPD = CD²

=> AB² = K²CD²

=> AB = K *CD

=> AB/CD = K

=> AP/CP = BP/DP = AB/CD = K

=> ΔABP~ΔCDP