Dadgarnia wrote: Point $P$ lies inside of parallelogram $ABCD$. Perpendicular lines to $PA,PB,PC$ and $PD$ through $A,B,C$ and $D$ construct convex quadrilateral $XYZT$. Prove that $S_{XYZT}\geq 2S_{ABCD}$. Official solution: Lemma. Point $A'$ is antipode of vertex $A$ in circumcircle of triangle $ABC$. prove that $$S_{ABC}-S_{A'BC}=\frac{1}{2}BC^2 \cot \angle A$$$proof.$ Let $H$ be the orthocenter of $ABC$. We have $$S_{ABC}-S_{A'BC}=S_{ABC}-S_{HBC}=\frac{1}{2}BC\cdot AH=\frac{1}{2}BC^2 \cot \angle A$$Now let $AB=a, \, AD=b, \, \angle APB=\angle P_1, \, \angle BPC=\angle P_2, \, \angle CPD=\angle P_3$ and $\angle DPA=\angle P_4$. From lemma we have \begin{align*} 2S_{ABCD}-S_{XYZT} &=\frac{a^2}{2}\left(\cot \angle P_1 + \cot \angle P_3\right)+\frac{b^2}{2}\left(\cot \angle P_2 + \cot \angle P_4\right) \\ &=\frac{a^2}{2}\left(\frac{\sin \left(\angle P_1+\angle P_3\right)}{\sin \angle P_1 \cdot \sin \angle P_3}\right)+\frac{b^2}{2}\left(\frac{\sin \left(\angle P_2+\angle P_4\right)}{\sin \angle P_2 \cdot \sin \angle P_4}\right) \end{align*}WLOG $\angle P_1+\angle P_3>180^\circ$ (If $\angle P_1+\angle P_3=180^\circ$ the result follows) so $\sin\left(\angle P_1+\angle P_3\right)<0$. We shall to prove that (Points $K,L,M$ and $N$ are the foot of the perpendicular lines from $P$ to $AB,BC,CD$ and $DA$ respectively) \begin{align*} &\Longleftrightarrow\frac{a^2}{2}\left(\frac{\sin \left(\angle P_1+\angle P_3\right)}{\sin \angle P_1 \cdot \sin \angle P_3}\right)+\frac{b^2}{2}\left(\frac{\sin \left(\angle P_2+\angle P_4\right)}{\sin \angle P_2 \cdot \sin \angle P_4}\right) < 0 \\ &\Longleftrightarrow \frac{a^2}{\sin \angle P_1 \cdot \sin \angle P_3} >\frac{b^2}{\sin \angle P_2 \cdot \sin \angle P_4} \\ &\Longleftrightarrow \frac{AP}{\sin \angle ABP}\cdot \frac{CP}{\sin \angle CDP} > \frac{AP}{\sin \angle ADP}\cdot \frac{CP}{\sin \angle CBP} \\ &\Longleftrightarrow \sin \angle CBP \cdot \sin \angle ADP > \sin \angle ABP \cdot \sin \angle CDP \\ &\Longleftrightarrow BP\cdot \sin \angle CBP \cdot DP\cdot \sin \angle ADP > BP\cdot \sin \angle ABP \cdot DP\cdot \sin \angle CDP \\ &\Longleftrightarrow PL\cdot PN > PK \cdot PM \\ &\Longleftrightarrow \angle LKN+\angle LMN > \angle KLM+\angle KNM \\ &\Longleftrightarrow \angle APB+\angle CPD > \angle APD+\angle BPC \end{align*}The last assertion is obvious so the proof is complete.

Lemma (well known). Suppose point $P$ is inside the circumcircle of triangle $ABC$, then the area of the pedal triangle of $P$ wrt $\triangle ABC$ (denoted by $\triangle DEF$) is given by $$S_{\triangle DEF}=\frac14(R^2-OP^2)\cdot \frac{S_{\triangle ABC}}{R^2}$$which, in particular, implies that $$S_{\triangle DEF}\le \frac14S_{\triangle ABC}.$$In the problem, suppose for example that $TX\perp PA$, $XY\perp PB$ and so on. Drop a perpendicular line from $P$ to $XZ$ at $E$. Then \begin{align*} S_{ABCD} &=2(S_{AED}+S_{BEC})\\ &\le 2(\frac14 S_{XTZ}+ \frac14 S_{XYZ})\\ &=\frac12 S_{XYZT}. \end{align*}There are two remaining configuration issues (which turns out to be quite messy): we need to prove that $P$ lies inside both the circumcircle of $XTZ$ and $XYZ$ (in order for the lemma to work), and also $E$ must lie between lines $AD$ and $BC$ (in order for the first equality above to work). To avoid other configuration issues, we use directed angles. One thing to notice, however, is that in terms of directed angles, the notion of an angle being larger than the other is a bit obscure since all of them are taken modulo $180^{\circ}$. Hence, we say directed angles $\alpha \ge \beta$ if the relation is true when they are in the interval $[0,180^{\circ})$. For the first issue, we can prove that as long as $P$ is inside $ABCD$, then $P$ is inside the circumcircle of, say, $XTZ$. There are two different cases: if $P,X,Z$ are located counterclockwise, we need $\measuredangle ZPX\le \measuredangle ZTX$, while if $P,X,Z$ are located clockwise, we need $\measuredangle ZPX\ge \measuredangle ZTX$. In the first case, \begin{align*} \measuredangle ZPX &=\measuredangle ZPD + \measuredangle DPA + \measuredangle APX\\ &=\measuredangle ZCD + \measuredangle DTA + \measuredangle ABX\\ &=\measuredangle ZTX + \measuredangle CPB \end{align*}Here, since we know $\measuredangle CPB \le 180^{\circ}$, we only need to prove that $\measuredangle ZTX + \measuredangle CPB$ is numerically $> 180^{\circ}$ (thus permitting it to be smaller than $\measuredangle ZTX$ when taken modulo $180^{\circ}$). This is equivalent to $\measuredangle CPB\ge \measuredangle APD$. \begin{align*} \measuredangle CPB &= \measuredangle AED\\ &= \measuredangle EAP + \measuredangle APD + \measuredangle PDE\\ &= \measuredangle EXP + \measuredangle APD + \measuredangle PZE\\ &\ge \measuredangle APD \end{align*}because of the formerly assumed orientation ($P,X,Z$ are located counterclockwise). (The first equality above follows from basic angle-chasing.) The second case (when $P,X,Z$ are located clockwise) follows basically the same steps. This concludes the first configuration issue. For the second issue, to characterize the condition "$D,A,E$ are clockwise", we can compute $\measuredangle DAE + \measuredangle EDA$ (added numerically, not modulo $180^{\circ}$) and check if the sum is $<180^{\circ}$. \begin{align*} \measuredangle DAE + \measuredangle EDA &= \measuredangle DAB + \measuredangle BAE + \measuredangle EDC + \measuredangle CDA\\ &= \measuredangle DAB - \measuredangle EXB - \measuredangle CZE + \measuredangle CDA\\ &= \measuredangle DAB + \measuredangle CDA + \measuredangle XYZ - 180^{\circ}\\ &= \measuredangle XYZ\\ &\le 180^{\circ} \end{align*}which means that $D,A,E$ are located clockwise. Similarly $B,C,E$ are clockwise, so $P$ is between $AD$ and $BC$. This concludes the second configuration issue, and thus completes our proof. Remark. I think these configuration issues are very messy, but I do not know of any way to avoid them except bashing the whole problem (as in the last post).