Let \(K =\) area of \(ABC, [\:] =\) the area of a polygon, \(b = AC,\) and \(c = AB.\) Note that, since \(AXYZ\) is a rhombus, \(AY\) is an angle bisector of \(\angle A.\) From the angle bisector theorem, we get \(\dfrac{BY}{YC} = \dfrac{c}{b}.\) From this, we get \(\dfrac{BY}{BC} = \dfrac{c}{b+c}\) and \(\dfrac{YC}{BC} = \dfrac{b}{b+c}.\)

\[\frac{[ZYC]}{K} = \left(\frac{b}{b+c}\right)^2 \Rightarrow [ZYC] = K \left(\frac{b}{b+c}\right)^2\]
\[\frac{[XYB]}{K} = \left(\frac{c}{b+c}\right)^2 \Rightarrow [XYB] = K \left(\frac{c}{b+c}\right)^2.\]

Now, \([AXYZ] = K - [ZYC] - [XYB] = K - K \left(\dfrac{b}{b+c}\right)^2 - K \left(\dfrac{c}{b+c}\right)^2 = K \left(\dfrac{2bc}{(b+c)^2}\right).\) Finally, we are left to prove that \(\dfrac{2bc}{(b+c)^2} \leq \dfrac{1}{2}.\) Rearranging this gives us \((b - c)^2 \geq 0,\) which is clearly always true. Thus, \([AXYZ] \leq \dfrac{1}{2}K,\) and our proof is complete.

Let ABCD be a rhombus such that A: (-a,0), B; (0,b), C: (a,0) & D: (0,-b) with centre of the rhombus as (0,0).
Now draw a line: y=mx+c passing through D and intersecting extended BA & BC in P & Q respectively.
We now have a triangle PQB with a rhombus whose vertices are on the triangle sides.
Since y=mx+c passes through D, c = -b. We can now determine P as intersection of y=mx -b & x/a+y/b=1 and thus, P:[2ab/(am-b), b(am+b)/(am-b)]. Similarly, Q:[2ab/(am+b), b(am-b)/(am+b)]. Further, we determine the area of triangle PQB taking the vertices in anti-clockwise direction.
Tr. PQB = (1/2){(2ab/(am-b))[b(am-b)/(am+b) -b)+(2ab/(am+b))[b -b(am-b)/(am-b)]}
This simplifies to, Tr. PQB = 4ab³/(b²-a²m² ) whereas atra pf Rhombus ABCD = 2ab.
Note that if m=0 then (1/2) Tr.PQB = 2ab = Area of Rhombus ABCD.
Otherwise (1/2)*4ab³/(b²-a²m² ) > 2ab if b²>b² - a²m² or if a²m²>0 which is always true.
Hence the proposition stands proven.

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## Comments

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TopNewestHere's an algebraic proof:

Let \(K =\) area of \(ABC, [\:] =\) the area of a polygon, \(b = AC,\) and \(c = AB.\) Note that, since \(AXYZ\) is a rhombus, \(AY\) is an angle bisector of \(\angle A.\) From the angle bisector theorem, we get \(\dfrac{BY}{YC} = \dfrac{c}{b}.\) From this, we get \(\dfrac{BY}{BC} = \dfrac{c}{b+c}\) and \(\dfrac{YC}{BC} = \dfrac{b}{b+c}.\)

\[\frac{[ZYC]}{K} = \left(\frac{b}{b+c}\right)^2 \Rightarrow [ZYC] = K \left(\frac{b}{b+c}\right)^2\] \[\frac{[XYB]}{K} = \left(\frac{c}{b+c}\right)^2 \Rightarrow [XYB] = K \left(\frac{c}{b+c}\right)^2.\]

Now, \([AXYZ] = K - [ZYC] - [XYB] = K - K \left(\dfrac{b}{b+c}\right)^2 - K \left(\dfrac{c}{b+c}\right)^2 = K \left(\dfrac{2bc}{(b+c)^2}\right).\) Finally, we are left to prove that \(\dfrac{2bc}{(b+c)^2} \leq \dfrac{1}{2}.\) Rearranging this gives us \((b - c)^2 \geq 0,\) which is clearly always true. Thus, \([AXYZ] \leq \dfrac{1}{2}K,\) and our proof is complete.

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Thanks .But I think AY is bisector because AXYZ is rhombus,a diagonal is not a bisector in parallelogram.

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Oops, sorry about that! I've fixed my solution.

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Let ABCD be a rhombus such that A: (-a,0), B; (0,b), C: (a,0) & D: (0,-b) with centre of the rhombus as (0,0). Now draw a line: y=mx+c passing through D and intersecting extended BA & BC in P & Q respectively. We now have a triangle PQB with a rhombus whose vertices are on the triangle sides. Since y=mx+c passes through D, c = -b. We can now determine P as intersection of y=mx -b & x/a+y/b=1 and thus, P:[2ab/(am-b), b(am+b)/(am-b)]. Similarly, Q:[2ab/(am+b), b(am-b)/(am+b)]. Further, we determine the area of triangle PQB taking the vertices in anti-clockwise direction. Tr. PQB = (1/2){(2ab/(am-b))[b(am-b)/(am+b) -b)+(2ab/(am+b))[b -b(am-b)/(am-b)]} This simplifies to, Tr. PQB = 4ab³/(b²-a²m² ) whereas atra pf Rhombus ABCD = 2ab. Note that if m=0 then (1/2) Tr.PQB = 2ab = Area of Rhombus ABCD.

Otherwise (1/2)*4ab³/(b²-a²m² ) > 2ab if b²>b² - a²m² or if a²m²>0 which is always true.

Hence the proposition stands proven.

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Really thanks for the proof.But is there any prove without using coordinate geometry?

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