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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

2/3

step-by-step explanation: