Proof: 2 Marks
Steps: 2 Marks

In $\mathrm{Δ}ABC\text{Â}and\text{Â}\mathrm{Δ}BAD$,
AD = BC [Given] 
$\mathrm{âˆ}DAB=\mathrm{âˆ}CBA$   [Given]
$AB=BA$   [Common]
$∴\mathrm{Δ}ABCâ‰\dots \mathrm{Δ}BAD$ (By SAS congruence rule)
$⇒DB=AC$   [Corresponding parts of corresponding triangles] ...... (1)
In $\mathrm{Δ}ADC\text{Â}and\text{Â}\mathrm{Δ}BCD$
$AD=BC$    [Given]
$DB=AC$    [From (1)]
$DC=CD\left(Common\right)$
$\mathrm{Δ}ADCâ‰\dots \mathrm{Δ}BCD$  [By SSS congruence rule]
$⇒\mathrm{âˆ}ACD=\mathrm{âˆ}BDC$  [Corresponding parts of corresponding triangles]