# A plane geometry tough problem

$ABCD$ is a quadrilateral. $P,Q,R,S$ are the midpoints of $AB,BC,CD,DA$ respectively. $PR$ and $SQ$ intersect at $L$. $T$ is any point within the quadrilateral. Prove that $4LT^2+LA^2+LB^2+LC^2+LD^2=TA^2+TB^2+TC^2+TD^2$.

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What have you tried so far? This is a pretty standard result. –  Calvin Lin Aug 3 at 4:26

Hint: $L$ is the midpoint of $PR$ (and also the midpoint of $QS$).

Observe that $RS = \frac{1}{2} AC = PQ$. Hence triangles $LRS, LPQ$ are congruent, so $LP=LR, LQ=LS$.

Hint: Apollonius' Theorem, which is the midpoint case of Stewart's Theorem.

Apply this many times, to the numerous triangles that we have. Specifically, $LAB, LCD, TAB, TCD, TPR, TQS$.

\begin{align} LAB: & LA^2 + LB^2 & = 2 LP^2 + 2 AP^2 \\ LCD: & LC^2 + LD^2 & = 2LR^2 + 2RC^2 \\ TAB: &TA^2 + TB^2 & = 2 TP^2 + 2 AP^2\\ TCD: &TC^2 + TD^2 &= 2 TR^2 + 2RC^2 \\ TPR: & 2TP^2 + 2TR^2 & = 4TL^2 + 4LR^2 \\ \end{align}

Hence $$\begin{array} { l l } TA^2 + TB^2 + TC^2 + TD^2 & = 2 TP^2 + 2 TR^2 + 2 AP^2 + 2 RC^2 \\ & = 4 TL^2 + 2LP^2 + 2AP^2 + 2LR^2 + 2RC^2 \\ & = 4 TL^2 + LA^2 + LB^2 + LC^2 + LD^2. \end{array}$$

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@sangabattulalokesh If you like the answer, you should accept it. –  Calvin Lin Aug 3 at 14:16
yeah i accept it is a good answer –  sangabattulalokesh Aug 9 at 16:37

Let the coordinates of the points $A$, $B$, $C$, and $D$ be, respectively, $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$, and $(x_4, y_4)$. Then the coordinates of the points $P$, $Q$, $R$, and $S$ are, respectively, $(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$, $(\frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2})$, $(\frac{x_3 + x_4}{2}, \frac{y_3 + y_4}{2})$, and $(\frac{x_4 + x_1}{2}, \frac{y_4 + y_1}{2})$.

The equation of the line through the points $P$ and $R$ is $$\frac{y- \frac{y_1 + y_2}{2}}{x- \frac{x_1 + x_2}{2}} = \frac{ \frac{y_3 + y_4}{2} - \frac{y_1 + y_2}{2} }{ \frac{x_3 + x_4}{2} - \frac{x_1 + x_2}{2} },$$ and similarly for the line through the points $Q$ and $S$.

Suppose the point $L$ has coordinates $(a,b)$, then this ordered pair satisfies both the equations of the lines through $P$ and $R$ and the line through $Q$ and $S$; so we can plug $a$ for $x$ and $b$ for $y$ in these two equations and solve the simultaneous equation to obtain the values of $a$ and $b$ in terms of $x_i$, $y_j$, where $i$, $j$ $= 1, 2, 3, 4$.

Now since the point $T$ lies in the quadrilateral $ABCD$, if we assume $(c,d)$ to be the coordinates of $T$, then we must have $$\min\{x_1, x_2, x_3, x_4 \} \leq c \leq \max\{x_1, x_2, x_3, x_4 \},$$ and $$\min\{y_1, y_2, y_3, y_4 \} \leq d \leq \max\{y_1, y_2, y_3, y_4 \}.$$

Hope this much hint will help.

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Note that position of point $T$ is irrelevant. It does not need to be within the quadrilateral. –  Calvin Lin Aug 3 at 14:15