Proposed Problem

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Complete Problem 324

Level: High School, SAT Prep, College geometry

## Saturday, July 18, 2009

### Problem 324: Quadrilateral, Perpendicular diagonals, Concurrence

Labels:
concurrent,
diagonal,
perpendicular,
point,
quadrilateral

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We first assume that FG & AC when extended meet in O and then prove that O, H & M are collinear.

ReplyDeleteFrom ECB, EC^2/CB=CG & EB^2/CB = GB or CG/GB = EC^2/EB^2. Likewise from Tr. ABE, BF/FA = EB^2/AE^2. Hence, CG/GB * BF/FA = EC^2/AE^2 - (1)

Now consider triangle ABC with FGO as the transversal. We can say, by Menelaus’s Theorem, that: CG/GB * BF/FA * AO/CO = -1 or CG/GB * BF/FA = - CO/AO.

Using eqn. (1), - CO/AO = EC^2/AE^2 or CO/AO = - EC^2/AE^2 ---------(2).

From triangles AED and DEC we can obtain: AM/MD = AE^2/ED^2 and DH/HC = ED^2/EC^2 or AM/MD * DH/HC = AE^2/EC^2.

Further, AM/MD * DH/HC * CO/AO = AE^2/EC^2 * - EC^2/AE^2 using eqn. (2)

In other words, AM/MD * DH/HC * CO/AO = -1 or by the converse of Menelaus’s Theorem, O, H & M are collinear which, in effect, means that FG, AC and MH are concurruent at O.

Alternatively, one may take EC & EB as the x-axis & y-axis resply. with A as (-a,0), B:(0,b), C:(c,0) and d:(0,-d). Writing eqns. of AB, BC, CD,DA,EG,EF,EM & EH one can obtain co-ordinates of G,F,M and H (Use QuickMath if you like since th algebra is a bit cumbersome) and finally equations of FG and MH to prove that they actually intersect on the x-axis at (ac/(a-c),0) which, in effect, means that the three concerned lines are, indeed, congruent.

Ajit: ajitathle@gmail.com

Erratum:

ReplyDeleteIn the proof above the last word must read "concurrent".

Ajit

Consider triangle ABC and points F and G

ReplyDeleteNote that B,E,F and G and cocyclic so ∠ (BEF)= ∠ (BGF)= ∠ (BAE)

So A,F,G and C are cocyclic and OC.OA=OG.OF=OE^2

On OE extension take a point E’ such that OE’=OE

So we have OE^2=OE’^2=OC.OA => E and E’ are points of conjugal harmonic of C and A

And point O is depend on positions of A,E and C only.

Similarly with triangle ADC and points M and H . the point of intersection of MH to AC is depend only on positions of A,E and C only

So AC, FG and MH are concurrent at one point O