2019-I-18

1. By applying sine law to $\Delta ABD$, we have

   $\begin{array}{rcl}
   \dfrac{BD}{\sin \angle BAD} & = & \dfrac{AD}{\sin \angle ABD} \\
   \dfrac{12}{\sin \angle BAD} & = & \dfrac{13}{\sin 72^\circ} \\
   \angle BAD & = & 61.389\ 869\ 36^\circ \\
   \end{array}$

2. By applying the sine law to $\Delta ABD$, we have

   $\begin{array}{rcl}
   \dfrac{AB}{\sin \angle ADB} & = & \dfrac{AD}{\sin \angle ABD} \\
   \dfrac{AB}{\sin (180^\circ – 72^\circ – 61.389\ 869\ 36^\circ)} & = & \dfrac{13}{\sin 72^\circ} \\
   AB & = & 9.933\ 216\ 094 \text{ cm}
   \end{array}$

   Note that $BP \perp AD$. Then in $\Delta ABP$,

   $\begin{array}{rcl}
   \cos angle BAD & = & \dfrac{AP}{AB} \\
   AP & = & 9.933\ 216\ 904 \times \cos 61.389\ 869\ 36^\circ \\
   AP & = & 4.756\ 491\ 614\text{ cm}
   \end{array}$

   Since $AC = AD = CD$, then $\Delta ACD$ is an equilateral triangle. Therefore, $\angle CAD = 60^\circ$.

   By applying cosine law to $\Delta ACP$, we have

   $\begin{array}{rcl}
   CP^2 & = & AC^2 + AP^2 – 2\times AC \times AP \cos \angle CAP \\
   CP^2 & = & 13^2 + 4.756\ 491\ 614^2 – 2 \times 13 \times 4.756\ 491\ 614 \times \cos 60^\circ \\
   CP & = & 11.392\ 533\ 59 \text{ cm}
   \end{array}$

$\begin{array}{rcl}
\cos \angle APC & = & \dfrac{CP^2 + AP^2 – AC^2}{2 \times CP \times AP} \\
\cos \angle APC & = & \dfrac{11.392\ 533\ 59^2 + 4.756\ 491\ 614^2 – 13^2}{2 \times 11.392\ 533\ 59 \times 4.756\ 491\ 616} \\
\angle APC & = & 98.803\ 115\ 12^\circ \\
& \neq & 90^\circ
\end{array}$

Therefore, $\angle BPC$ is not the angle between the face $ABD$ and the face $ACD$.

Hence, the claim is not correct.