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# Seeking Help

In triangle find points X,Y,Z on AB,BC,CA such that AXYZ is a rhombus.Show that the area of rhombus AXYZ≤(1/2) triangle ABC. Please show me the proof

Note by Kalpok Guha
3 years, 3 months ago

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Here'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.

- 3 years, 3 months ago

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

- 3 years, 3 months ago

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

- 3 years, 3 months ago

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.

- 3 years, 3 months ago

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

- 3 years, 3 months ago