Line segments AB and CD bisect each other at P. If PA = PD and PB = PC, prove that AC = BD.
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AB and CD bisect each other at O i.e, AO=BO and CO=DO
in ΔCOA and ΔDOB
Given CO=OD,∠COA=∠BOD [ vertically opp angles]
AD=BD
∴ΔCOA≅ΔBOD
(i) ∴AC=BD[C.P.CT]
(ii) ∠CAB=∠ABD[C.P.CT]
again
in ΔCOB and ΔAOD
CO=OD [given]
BO=AO [given]
∠COB=∠AOD [vertically opp angles]
∴ΔCOB≅ΔAOD
∴∠CBA=∠BAD [ C.P. C.T]
(iii) and so AD∣∣CD [ ∵∠CBA=∠BAD which are altanate angles]
and AD=CB [C.P.C.T]
Step-by-step explanation:
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