Let-ABCD-be-a-convex-quadrilateral-Prove-that-the-orthocenters-of-the-triangles-ABC-BCD-CDA-and-DAB-are-the-vertices-of-a-quadrilateral-congruent-to-ABCD-and-prove-that-the-centroids-of-the-same-tr

Question Number 16072 by Tinkutara last updated on 17/Jun/17

Let A B C D be a convex quadrilateral .

Prove that the orthocenters of the

triangles A B C, B C D, C D A and D A B

are the vertices of a quadrilateral

congruent to A B C D and prove that the

centroids of the same triangles are the

vertices of a cyclic quadrilateral .

Commented by mrW1 last updated on 06/Jul/17

the question seems to be wrong . I think

the centroids are the vertices of a

quadrilateral congruent to ABCD .

For part 1 please post a figure to

show what is meant .

Commented by mrW1 last updated on 06/Jul/17

yes please .

Commented by mrW1 last updated on 06/Jul/17

I can prove that the centroids are the vertices of a

quadrilateral congruent to ABCD . Its

area is \frac{1}{9} of the area of ABCD .

Answered by Tinkutara last updated on 07/Jul/17

Commented by Tinkutara last updated on 07/Jul/17

Lemma used :

\vec{O A} + \vec{O B} + \vec{O C} = \vec{O H} and

\vec{O H} = 3 \vec{O G} .

Commented by Tinkutara last updated on 07/Jul/17

Proof :

Let O be circumcenter of the

quadrilateral A B C D and let H_{A}, H_{B},

H_{C}, H_{D} be the orthocenters of the

triangles B C D, C D A, D A B and

A B C, respectively . Using the lemma,

we have

\vec{H_{A} H_{B}} = \vec{{O H}_{B}} - \vec{{O H}_{A}}

= (\vec{O C} + \vec{O B} + \vec{O A}) - (\vec{O B} + \vec{O C} + \vec{O D})

= \vec{O A} - \vec{O B} = \vec{B A},

and consequently the segments H_{A} H_{B}

and A B are parallel and equal in

length . We conclude that the

quadrilaterals A B C D and H_{A} H_{B} H_{C} H_{D}

are congruent .

For the second claim, let G_{A}, G_{B}, G_{C}

and G_{D} denote the centroids . It follows

from the observation above that

G_{A} G_{B} G_{C} G_{D} is obtained from

H_{A} H_{B} H_{C} H_{D} by a homothetic of center

O and ratio \frac{1}{3} . Because H_{A} H_{B} H_{C} H_{D}

is cyclic, the same is true of G_{A} G_{B} G_{C} G_{D} .

(See Figure 5.27) .

Commented by Tinkutara last updated on 07/Jul/17

Commented by mrW1 last updated on 07/Jul/17

the question {didn}^{'} t say that ABCD is

cyclic .

Commented by mrW1 last updated on 07/Jul/17

the answer given {doesn}^{'} t match the

question . i can not understand .

Commented by mrW1 last updated on 07/Jul/17

i have said the question is not correct .

so i can not give a solution to it .

the proof i meant is following . but it

is not the proof to your question exactly .

Commented by mrW1 last updated on 07/Jul/17

Commented by mrW1 last updated on 07/Jul/17

ABCD is an any convex quadrilateral .

K, L, M, N are the centroids described .

\frac{GM}{GB} = \frac{1}{3}

\frac{GN}{GC} = \frac{1}{3}

\Rightarrow MN / / BC and MN = \frac{BC}{3}

similarly

NL / / AB and NL = \frac{AB}{3}

LK / / AD and LK = \frac{AD}{3}

KM / / DC and KM = \frac{DC}{3}

\Rightarrow LKNM is congruent to ABCD

\Rightarrow Area [KLNM] = \frac{1}{9} Area [ABCD]

Commented by mrW1 last updated on 07/Jul/17

this proof is better than that given in

your book, i think .

ABCD must not be cyclic!

Commented by Tinkutara last updated on 08/Jul/17

Thanks Sir!