In right △ABC,∠C=900,M is the mid-point of hypotenuse A-B. C is jointed to M and produced to D such that DM= CM. Point D is jointed to B. Show that △DBC≅△ACB.
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Solution
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In△BDM&△AMC
BM=CM(given)
∠DMB=∠AMC(verticallyopposite)
BM=AM(given)
∴△BDM≅△AMCbyS.A.Scongruency
∴AC=DB(C.P.C.T)−(1)
and∠ACM=∠BDM(C.P.C.T)
and∠CAM=∠DBM(C.P.C.T)
∴AC||BD(∵alternateanglesequal)
∴∠ACB=∠DBC(correspondingangles)
(∵AC||BDandBCisthecommontransverse)
∴∠DBC=90∘
In△DBC&△ACB
BC=CB(common)
∠DBC=∠ACB=90∘(proved)
DB=AC(from1)
∴DBC≅△ACB
Hence,proved
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