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?
Best Answer
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?