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Perpendicularity
April   30
N 2 minutes ago by Tsikaloudakis
Source: CGMO 2007 P5
Point $D$ lies inside triangle $ABC$ such that $\angle DAC = \angle DCA = 30^{\circ}$ and $\angle DBA = 60^{\circ}$. Point $E$ is the midpoint of segment $BC$. Point $F$ lies on segment $AC$ with $AF = 2FC$. Prove that $DE \perp EF$.
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Diagonals of a Trapezium
DumSpiroSpero   2
N Jul 10, 2016 by Virgil Nicula
Source: http://mathworld.wolfram.com/Trapezoid.html
In a trapezium with bases $a$ and $b$ and lateral sides $c$ and $d$, proof that the diagonals $p$ and $q$ are given by the following expressions, respectively:

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DumSpiroSpero
Jul 9, 2016
Virgil Nicula
Jul 10, 2016
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Diagonals of a Trapezium
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Source: http://mathworld.wolfram.com/Trapezoid.html
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DumSpiroSpero
68 posts
DumSpiroSpero
#1 h  Jul 9, 2016, 3:13 PM • 1 Y
Y by Adventure10
In a trapezium with bases $a$ and $b$ and lateral sides $c$ and $d$, proof that the diagonals $p$ and $q$ are given by the following expressions, respectively:

http://i.imgur.com/DtAZitz.gif

http://i.imgur.com/Z6qQDAn.gif
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george_54
1585 posts
george_54
#2 h  Jul 9, 2016, 4:50 PM • 2 Y
Y by Adventure10, Mango247
Applying Euler's theorem:
${a^2} + {b^2} + {c^2} + {d^2} = {p^2} + {q^2} + 4M{N^2} = {p^2} + {q^2} + 4\frac{{{{(b - a)}^2}}}{4} \Leftrightarrow $ $\boxed{{p^2} + {q^2} = {c^2} + {d^2} + 2ab}$ $(1)$

Cosine Law:
$\left\{ \begin{array}{l} {p^2} = {b^2} + {c^2} - 2bc\cos D\\ {q^2} = {a^2} + {c^2} - 2ac\cos A \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} a{p^2} = a{b^2} + a{c^2} + 2abc\cos A\\ b{q^2} = {a^2}b + b{c^2} - 2abc\cos A \end{array} \right. \Leftrightarrow $ $\boxed{a{p^2} + b{q^2} = a{b^2} + a{c^2} + {a^2}b + b{c^2}}$ $(2)$

From $(1), (2)$ we get the desired result.
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This post has been edited 1 time. Last edited by george_54, Jul 9, 2016, 4:59 PM
Reason: typo
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Virgil Nicula
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Virgil Nicula
#3 h  Jul 10, 2016, 2:20 AM • 2 Y
Y by DumSpiroSpero, Adventure10
See P1 from here
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