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In the convex quadrilateral $ABCD$, which is not a parallelogram, the line passing through the centers of the diagonals $AC$ and $BD$ intersects the segment $BC$ at $R$. How to prove that the sum of the areas of triangles $ABR$ and $CDR$ is equal to the area of triangle $ADR$? I have no idea how to do this. Can this be proved with simple geometry?

enter image description here

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  • $\begingroup$ Is it a true property this time ? Do you see what I mean ? $\endgroup$ – Jean Marie Mar 15 '16 at 18:32
  • $\begingroup$ I think that $ABCD$ is not a kite either. $\endgroup$ – mrprottolo Mar 15 '16 at 18:40
  • $\begingroup$ @maxkor I don't understand "the line passing through the centers of the diagonal intersects the segment $BC$ at $R$". $\endgroup$ – Weijie Chen Mar 17 '16 at 12:38
  • $\begingroup$ @WeijieChen I've attached the draw $\endgroup$ – piteer Mar 17 '16 at 17:50
  • $\begingroup$ This is a good question. It took me one week's time to figure out the solution. After it has been proven, there will be one more (a much simpler) way to find half the area of a quadrilateral. $\endgroup$ – Mick Mar 22 '16 at 19:01
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This is tricky.

We shall use an easy fact that if we are given fixed points $Y,Z$, and a variable point $X$ that changes linearly, then $[XYZ]$ changes linearly, where $[\mathcal{F}]$ denotes the oriented area of $\mathcal{F}$.

Let $M,N$ be midpoints of $AC, BD$. Using the fact we know that the function $MN \ni X \mapsto [ABX]+[CDX]$ is linear. However $$[ABN]+[CDN]=\frac 12 [ABD] + \frac 12 [CDB] = \frac 12 [ABCD]$$ and $$[ABM]+[CDM]=\frac12 [ABC] + \frac 12 [CDA] = \frac 12 [ABCD]$$ so this function is actually constant. In particular $$[ABR]+[CDR]=\frac 12 [ABCD].$$ This implies that $$[DAR] = [ABCD]-([ABR]+[CDR])=[ABCD] - \frac 12 [ABCD]=\frac 12[ABCD] = [ABR]+[CDR].$$

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Let H and K be the midpoints of AC and BD respectively.

To prove the required, I will break the proof into several claims.

Claim #1: Every quadrilateral can be bisected (by a straight line) into two halves that are equal in area.

After joining AK and CK, ABCD is now divided into 4 triangles with [red] = [pink] and [blue] = [green]. Then, [pink] + [blue] = [white] + [green]. (See figure 1.)

enter image description here

Construction: (1) Draw KE // AC cutting AB at E; and (2) Let CE cut AK at J. By “equal base, equal altitude” principle, [⊿AEJ] = [⊿CKJ]. This means we can transform ⊿AEJ to fit the position of ⊿CKJ such that [⊿BEC] = [quad AECD]. (See figure 3.)

enter image description here

After adding ⊿RAD, the next goal is to fill its area by pieces of quad AECD. ⊿RAD is partly filled by Quad APUD because it is common to both. Note that [⊿AEP] = [⊿TUC] + [quad QPUT]. The latter is used to cover part of ⊿RAD while ⊿TUC is to be connected to ⊿CUD to form the new ⊿TCD.

The remaining question is “will [⊿TCD] = [⊿RQT]?” We are done if we can prove DQ // BC.

Claim #2: BVDC is a parallelogram.

enter image description here

Further construction needed: (1) Extend CK to V such that AV // HK; (2) Join BV.

Applying intercept theorem to ⊿CAV, we have CK = KV. This and with the given BK = KD make BVDC a parallelogram.

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This link gives a great visual representation of this proof.
https://www.mathsisfun.com/geometry/quadrilaterals-interactive.html.

Analytically, I created a complicated proof by using the equation base*height/2 = area_of_a_triangle and cos(angle)*height = base for each of ADR & CDR & ADR and solved for a common relationship. I recommend finding another way.

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  • $\begingroup$ I have clicked on the link you provide, but one falls on a "naked" quadrilateral. $\endgroup$ – Jean Marie Mar 15 '16 at 22:33
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I am sorry, I withdraw (not completely ; see at the bottom) what I have said. I had read "line $AB$" instead of "segment $AB$".

Nevertheless, I think that, instead of "the line passing..." one should have found "in the case where the line passing through the midpoints $I$ et $J$ intersects segment $[BC]$, then ...".

I have looked for a geometric proof, but found none. Here is an analytical proof.

Take coordinate axis such that $B(-1,0)$ and $C(1,0)$ (without loss of generality). Let $A(a,b)$ and $D(c,d)$. Let $R(r,0)$.

The collinearity condition between $R$ and $I,J$ is

$$\begin{vmatrix}r&\frac{a+1}{2}&\frac{c-1}{2}\\ 0&\frac b2&\frac d2\\ 1&1&1\end{vmatrix}=0 \ \ \Rightarrow \ \ r=\dfrac{b+d+ad-bc}{2(d-b)} \ \ (1)$$

Now, it is a matter of (rather simple) computation to check that the difference of areas:

$$\delta=A(ABR)+A(CDR)-A(ADR)$$ is zero. Indeed,

$$\delta=\frac12det(\vec{RB},\vec{RA})+\frac12det(\vec{RC},\vec{RD})-\frac12det(\vec{RD},\vec{RA})$$

which is equivalent (after multiplication by $2$) to:

$$2\delta=\begin{vmatrix}a-r&-1-r\\b&0\end{vmatrix}+\begin{vmatrix}1-r&c-r\\0&d\end{vmatrix}+\begin{vmatrix}a-r&c-r\\b&d\end{vmatrix}$$

Expanding previous expression and replacing by the value of $r$ given by (1) gives $\delta=0$, as desired.

Note that a condition of existence of $R$ is $b \neq d$, i.e., trapezoids are not accepted (one can check indeed that, in such a case, line $IJ$ is parallel to $BC$).

Remark: The result, true for $R \in [BC]$, is no longer true if the line joining the midpoints of the diagonal intersect line BC outside segment [BC], if we continue to work with unsigned areas.

Here is a counterexample.

Let $A(0,2), B(1,0),C(2,0),D(3,1)$.

This gives for the midpoints: $I(1,1)$ and $J(2,1/2)$. Thus $R(3,0)$.

The areas are $A(ABR)=2, A(CDR)=1/2$ whereas $A(ADR)=3/2$ instead of $5/2$.

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  • $\begingroup$ According to your example, $R$ is not in the segment $BC$, unlike what OP asked. $\endgroup$ – Quang Hoang Mar 15 '16 at 19:44
  • $\begingroup$ I also tried something like that, but failed, I think this time the statement could be true. $\endgroup$ – mrprottolo Mar 15 '16 at 19:52

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