All faces of the tetrahedron $ABCD $ are acute-angled triangles.$AK$ and $AL$ -are altitudes in faces $ABC$ and $ABD$. Points $C,D,K,L$ lies on circle. Prove, that $AB \perp CD$
Problem
Source: St Petersburg Olympiad 2008, Grade 11, P5
Tags: geometry, 3D geometry, tetrahedron
Pathological
27.05.2019 02:56
Pretty easy for Russian 3-D Geo We wish to show that $AC^2 - AD^2 = BC^2 - BD^2.$ Let $P$ be the foot from $A$ onto the face $BCD$. Observe that $KP \perp BC$ and $LP \perp BD$, say by simple Pythagorean. Hence, we have that $\angle PBK = \angle PLK - 90 \angle BLK = 90 - \angle BCD$, where we used the cyclic condition. This implies that $BP \perp CD$, and so hence $AC^2 - AD^2 = CP^2 - DP^2 = BC^2 - BD^2$ by a few applications of Carnot, we're done. $\square$
PigeonBar
27.05.2019 05:46
As in Pathological's solution, let $P$ be the projection of $A$ onto $\Delta BCD$. Also, let $AM$ be an altitude of $\Delta ACD$.
Then, since $AP$ is perpendicular to the plane of $\Delta BCD$, $CD \perp AP$. Additionally, since $AM$ is an altitude of $\Delta ACD$, $CD \perp AM$. Thus, $CD$ is perpendicular to the plane of $\Delta APM$, so $PM \perp CD$. Similarly, $PK \perp BC$ and $PL \perp BD$.
Note that $K$, $L$, $M$, and $P$ are all in the same plane as $\Delta BCD$. We are given that quadrilateral $CDKL$ is cyclic. Quadrilaterals $CKPM$ and $DLPM$ are also cyclic, using the circles with diameters $CP$ and $DP$, respectively. Then, by the Radical Axis Theorem, lines $CK$, $DL$, and $PM$ are concurrent. Therefore, $B$ lies on $PM$, so since $PM \perp CD$, we get $BP \perp CD$. Finally, as mentioned above, $AP \perp CD$, so $CD$ is perpendicular to the plane of $\Delta ABP$.
$\therefore AB \perp CD$, as required.