ABC is an isosceles triangle in …

{tex}\Delta{/tex}ABD where AB = AC
AD bisects {tex}\angle{/tex}PAC,
& CD {tex}\|{/tex} AB
To prove : {tex}\angle{/tex}DAC = {tex}\angle{/tex}BCA
Proof:
AD bisects {tex}\angle{/tex}PAC
Hence {tex}\angle{/tex}PAD = {tex}\angle{/tex}DAC = {tex}\frac{1}{2}{/tex} {tex}\angle{/tex}PAC ...(i)
Also, given
AB = AC
{tex}\therefore{/tex} {tex}\angle{/tex}BCA = {tex}\angle{/tex}ABC (Angles opposite to equal sides are equal) ... (ii)
For {tex}\Delta{/tex}ABC,
{tex}\angle{/tex}PAC is an exterior angle
so, {tex}\angle \mathrm{PAC}=\angle \mathrm{ABC}+\angle \mathrm{BCA}{/tex} (Exterior angle is sum of interior opposite angles)
{tex}\angle \mathrm{PAC}=\angle \mathrm{BCA}+\angle \mathrm{BCA}{/tex} (From (2) : {tex}\angle \mathrm{ABC}=\angle \mathrm{BCA}{/tex})
{tex}\angle \mathrm{PAC}=2 \angle \mathrm{BCA}{/tex}
{tex}\frac{1}{2} \angle \mathrm{PAC}=\angle \mathrm{BCA}{/tex}
{tex}\angle \mathrm{BCA}=\frac{1}{2} \angle \mathrm{PAC}{/tex}
{tex}\angle B C A=\angle D A C \quad\left(\text { From }(1): \angle D A C=\frac{1}{2} \angle P A C\right){/tex}
Hence proved