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There's this geometry problem I tried to solve, but I am not sure if my solution is the easiest one. First I'll explain the problem:

Let's say we have a triangle ABC, a point X in the third of side AB closer to A, a point Y in the third of XB closer to X and then a point Z on BC placed in such a way so that the angles ∠XCB and ∠YZB are the same. What is the ratio of the areas of triangles ABC and XZC.

The way I solved this is by having a triangle ABC in which the height of the triangle passes through point X and its height is equal to XB. Because we know that triangles XCB and YZB share 1 angle and one of the same size (∠ABC and ∠XCB/∠YZB) we can, therefore, say that ∠BYZ=∠BXC=90°. Make a point H on XC, through which the height of XZC passes. That means that the angle ∠XHZ=90°. We can prove that the length of XY equals HZ, by looking at the fact that lines XH and YZ are parallel and because of the fact that the angle ∠XHZ=90° then also lines XY and HZ are parallel which makes XYZH a rectangle. So now we know that XY=HZ. We can now calculate the areas:

$$ \text{A of ABC} = {\dfrac{1\cdot\dfrac{2}{3}}{2}} = {\frac{1}{3}} $$ $$ \text{A of XZC} = {\dfrac{\dfrac{2}{3}\cdot{\dfrac{2}{9}}}{2}} = {\frac{2}{27}} $$ $$ \text{Ratio} = {\frac{1}{3}}\div{\frac{2}{27}} = 4.5 $$

Is there an easier way of calculating this?

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Lets say that triangle XZC has some area S.

Consider triangle BYZ and BXC. They are similar since they have shared angle B and angle BZY is equal to angle BCX, its also given that BY = 2/3 BX which, combined, implies that BZ = 2/3 * BC => BZ = 2 * ZC.

Now lets consider triangles XBZ and XZC. They have shared vertice X and their bases lay on a shared line BC, which implies that they have some shared height h. Area of XBZ is 1/2 * h * BZ and Area of XZC is 1/2 * h * ZC. But we know that BZ = 2 * ZC, so Area of XBZ is twice Area of XZC(which we stated to be "S", so 2*S).

Now lets consider triangles BXZ a BAC. They are similar since: 1)have shared angle ABC and 2)AB/XB = CB/ZB we also are given the fact that AB is 1.5 times bigger than XB so concluding that Area of ABC is 1.5^2=2.25 times bigger than Area of XBZ, which we know, is 2 times bigger than S. So we get Area of ABC is 2.25 * 2 * S = 4.5 * S QED?

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  • $\begingroup$ Please use MathJax formatting. $\endgroup$
    – TheSimpliFire
    Oct 12, 2019 at 9:26

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