269. Some nice and easy properties in a trapezoid.

by Virgil Nicula, Apr 25, 2011, 12:41 AM

PP1 (O.M. - Moscova, 2011). Let $ABCD$ be a trapezoid with $AD\parallel BC$ . Denote the midpoints $M$ and $N$ of

$[AB]$ and $[CD]$ respectively. Define the point $P$ for which $PM\perp AB$ , $PN\perp CD$ . Prove that $PA=PD$ .


Proof. Let midpoint $Q$ of $[AD]$ . Observe that $P$ is orthocenter of $\triangle MNQ$ . In conclusion, $PQ\perp MN$ , i.e. $PQ\perp AD$ $\iff PA=PD$ .


PP2. Let a trapezoid $ABCD$ , where $ AB\parallel CD$ . Denote $ O\in AC\cap BD$ and the midpoint $M$ of

side $[CD]$ . The circumcircles of $\triangle AOD$ , $\triangle BOC$ intersect at $K$ . Prove that $ \widehat{KOA} \equiv\widehat{MOD}$ .


Proof.

$ \blacktriangleright\ \left\|\begin{array}{ccc} \widehat {KAC}\equiv\widehat {KAO}\equiv\widehat {KDO}\equiv\widehat {KDB} & \implies & \widehat {KAC}\equiv\widehat {KDB} \\
 \\
\widehat {KCA}\equiv\widehat {KCO}\equiv\widehat {KBO}\equiv\widehat {KBD} & \implies & \widehat {KCA}\equiv\widehat {KBD} \end{array}\right\|$ $ \implies$ $ KAC\sim KDB$ $ \implies$ $ \frac {KA}{KD} = \frac {AC}{DB}$ $ \implies$ $ \boxed {\ \frac {KA}{KD} = \frac {OC}{OD}\ }\ (1)$ .

$ \blacktriangleright$ Denote $ S\in OK\cap CD$ . Apply Sinus' theorem in $ \triangle KAD\ \ :$ $ \frac {KA}{KD} = \frac {\sin\widehat {KDA}}{\sin\widehat {KAD}} = \frac {\sin\widehat {KOA}}{\sin\widehat {SOD}} = \frac {\sin\widehat {SOC}}{\sin\widehat {SOD}}$ , i.e. $ \boxed {\ \frac {KA}{KD} = \frac {\sin\widehat {SOC}}{\sin\widehat {SOD}}\ }\ (2)$ .

$ \blacktriangleright$ In $\triangle  COD$ apply an well-known property $ \boxed {\ \frac {SC}{SD} = \frac {OC}{OD}\cdot\frac {\sin\widehat {SOC}}{\sin\widehat {SOD}}\ }\ (3)$ . From $ (1)$ , $ (2)$ , $ (3)$ obtain $ \frac {SC}{SD} = \left(\frac {OC}{OD}\right)^2\implies$ $[OS$ is symmedian in $\triangle  COD$ .

Remark. If denote $ R\in OK\cap AB$ , then the ray $ [OR$ is symmedian in the triangle $ AOB$ .



PP3. $ABCD$ is a trapezoid with $AB\parallel CD$ . Let $P\in (AD)$ and $Q\in (BC)$ so that

$PQ\parallel AB$ and $[APQB]=[CDPQ]$ . Prove that $AB^2+CD^2 = 2\cdot PQ^2$ .


Proof. Let $AB=a\ ,\ CD=b\ ,\ PQ=r$ . Observe that $\frac {\delta_{AB}(D)}{\delta_{PQ}(D)}=\frac {DA}{DP}=\frac {a-b}{r-b}$ and $[APQB]=[CDPQ]\iff$

$\frac {[ABCD]}{[CDPQ]}=2\iff$ $\frac {a+b}{r+b}\cdot\frac {\delta_{AB}(D)}{\delta_{PQ}(D)}=2\iff$ $\frac {a+b}{r+b}\cdot\frac {a-b}{r-b}=2\iff$ $a^2-b^2=2\cdot \left(r^2-b^2\right)\iff$ $\boxed{a^2+b^2=2r^2}$ .

Remark. $\frac {PD}{PA}=\frac {a-b+2r}{a+b}$ , where $2r^2= a^2+b^2$ . Generally, $[CDPQ]=\rho\cdot [ABQP]\ \iff\ \rho\cdot a^2+b^2=(\rho +1)\cdot r^2$ .
This post has been edited 15 times. Last edited by Virgil Nicula, Nov 22, 2015, 8:33 AM

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