Answer:
Step-by-step explanation:
4) In ΔABD & ΔBAC
AD = BC {given}
∠DAB = ∠CBA {given}
AB = AB {Common to both triangles}
ΔABD ≅ ΔBAC { Side Angle Side - SAS congruent}
b) BD = AC { Corresponding Part of congruent trianlge}
5) (i)In ΔDMB & ΔAMC
M is midpoint of AB
MA = MB
∠DMB =∠AMC {Vertically opposite angles}
CM = DM {Given}
ΔDMB ≅ ΔAMC ( Side Angle Side - SAS congruent}
(ii) ∠DBM = ∠ACM {CPCT}
Alternate angles are equal. Therefore, BD // AC
BD // AC & BC transversal
Co interior angles are 180
∠DBC + ∠ACB = 180
∠DBC + 90 = 180
∠DBC = 180 - 90
∠DBC = 90 , ∠DBC is right angle.
(iii)In ΔDBC & ΔACB
BD = AC { CPCT form Proof (i)}
∠DBC = ∠ACB = 90 {from Proof (ii)}
BC = BC {common}
ΔDBC ≅ ΔACB { SAS Congruent}
(iv) CD = AB
C is midpoint of CD
CD = 2CM
2CM = AB
CM =(1/2)AB
