2019-I-18

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

   $\begin{array}{rcl}{\displaystyle \frac{BD}{\sin \mathrm{\angle }BAD}}& =& {\displaystyle \frac{AD}{\sin \mathrm{\angle }ABD}}\\ {\displaystyle \frac{12}{\sin \mathrm{\angle }BAD}}& =& {\displaystyle \frac{13}{\sin {72}^{\circ }}}\\ \mathrm{\angle }BAD& =& 61.389869{36}^{\circ }\end{array}$

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

   $\begin{array}{rcl}{\displaystyle \frac{AB}{\sin \mathrm{\angle }ADB}}& =& {\displaystyle \frac{AD}{\sin \mathrm{\angle }ABD}}\\ {\displaystyle \frac{AB}{\sin ({180}^{\circ }–{72}^{\circ }–61.389869{36}^{\circ })}}& =& {\displaystyle \frac{13}{\sin {72}^{\circ }}}\\ AB& =& 9.933216094\text{ cm}\end{array}$

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

   $\begin{array}{rcl}\cos angleBAD& =& {\displaystyle \frac{AP}{AB}}\\ AP& =& 9.933216904\times \cos 61.389869{36}^{\circ }\\ AP& =& 4.756491614\text{ cm}\end{array}$

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

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

   $\begin{array}{rcl}C{P}^2& =& A{C}^2+A{P}^2–2\times AC\times AP\cos \mathrm{\angle }CAP\\ C{P}^2& =& {13}^2+4.756491{614}^2–2\times 13\times 4.756491614\times \cos {60}^{\circ }\\ CP& =& 11.39253359\text{ cm}\end{array}$

$\begin{array}{rcl}\cos \mathrm{\angle }APC& =& {\displaystyle \frac{C{P}^2+A{P}^2–A{C}^2}{2\times CP\times AP}}\\ \cos \mathrm{\angle }APC& =& {\displaystyle \frac{11.392533{59}^2+4.756491{614}^2–{13}^2}{2\times 11.39253359\times 4.756491616}}\\ \mathrm{\angle }APC& =& 98.803115{12}^{\circ }\\ & \ne & {90}^{\circ }\end{array}$

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

Hence, the claim is not correct.