Given : Right triangle ABC, right angle is at C, M is mid- point of hypotenuse AB and DM=CM
To show: ∠DBC is a right angle
Proof :
in triangle DBM and triangle AMC,
DM=MC ( Given )
BM=MA ( Given )
∠DMB=∠AMC (Vertically opposite angle)
Therefore, by SAS criterion of congruency, triangle DMB is congruent to triangle AMC So,
DB=AC (By CPCT)
∠BDM= ∠CAM (by CPCT)
Now, in triangle DBC and triangle ABC,
DB=AC ( proved above)
BC=BC ( common)
DM=MA,MC=MB ( proved above)
so DM+MC=MA+MB
DC=AB
Therefore, by SSS criterion of congruency, triangle DBC is congruent to triangle ACB.
Therefore,
∠DBC=∠ACB=90∘ (By CPCT)