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Here is a triangle, whose area is $\frac{1}{2}(V-V_o)t$, where $V_o$ and $V$ are $y$-coordinates.

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$\frac{1}{2}(V-V_o)t$ is also $\frac{1}{2}Vt - \frac{1}{2}V_ot$ i.e., difference between two other triangles.

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While this is true, how to geometrically see how this works?

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    $\begingroup$ The triangle in blue and the triangle with the dashed sides have the same altitude and the same base. $\endgroup$ – Bernard Oct 30 '19 at 19:32
  • $\begingroup$ this may help $\endgroup$ – AgentS Oct 30 '19 at 19:33
  • $\begingroup$ Because the area of a triangle is base times the height and divide by 2. The base is in both cases $V-V_0$ and the height is in both cases $t$: $A=\frac{(V-V_0)\cdot t}{2}$ $\endgroup$ – callculus Oct 30 '19 at 19:33
  • $\begingroup$ mathworld.wolfram.com/CavalierisPrinciple.html $\endgroup$ – Xander Henderson Oct 30 '19 at 19:35
  • $\begingroup$ thanks all. Didn't know to see this way. And those links and you guys are very helpful. $\endgroup$ – Saran Oct 30 '19 at 19:37
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The key is first notice every triangle can produce a parallelogram by joining a copy of itself in inverse direction. Now it suffice to show their parallelograms have equal area. For parallelogram with inner altitude you can see it can be rectangle with equal area by replacing the right triangle that altitude separate it. In case of outer altitude you can check yourself as an exercise.

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