In Fig., D is a point on hypotenuse AC of ΔABC, such that BD⊥AC, DM⊥BC and DN⊥AB. Prove that :(i) DM²=DN.MC (ii) DN²=DM.AN

i) InΔABC,DN⊥ABandBC⊥ABSo,DN∥BC.. (1)DM⊥BCandAB⊥BCSo,DM∥AB.. (2)From (1) and (2),□DMBNis a rectangle.∴BM=DNInΔBMD,∠M+∠BDM+∠DBM=180o⇒∠BDM+∠DBM=90o.. (1)Similarly, inΔDMC,∠CDM+∠MCD=90o.. (2)We know,BD⊥ACgiven∴∠BDM+∠MDC=90o. (3)From (1) and (3), we get∠BDM+∠DBM=∠BDM+∠MDC∴∠DBM=∠MDC.. (4)Similarly,∠BDM=∠MCD.. (5)InΔBMDandΔDMC,∠BMD=∠DMC...Each90o∠DBM=∠MDC...From (4)∠BDM=∠MCD...From (5)ΔBMD∼ΔDMCAAA test of similarity∴BMDM=MDMCC.S.S.T.∴DNDM=DMMC...∵BM=ND⇒DM2=DN×MCii) Similarly, we can proveΔDNB∼ΔDNABNDN=NDNADMDN=DNAN...[∵BN=DM]DN2=DM×AN