This definition makes sense to me, and intuitively it seems like the larger $\alpha$ the more A is weighted. I was just wondering why this works mathematically (e.g. why must $\alpha + \beta + \gamma = 1$)?

go-lauren

@julialuo Call the point in question P. The line they drew in the picture separated by (1 - alpha) and alpha is the altitude of the triangle ABC from the vertex A. We can see the length alpha is the altitude of the triangle PBC. Thus, the area of ABC is 1/2 * b * h or 1/2 * BC * 1 and the area of PBC is 1/2 * BC * alpha, the the proportion of area of PBC to ABC or PBC/ABC is alpha.

We can repeat this comparison of altitudes for the altitudes of ABC drawn from B and C, so we get that beta and gamma are the proportions of the area of PAC and PAB to the entire triangle, respectively. Visually, we can easily see that PAC and PAB and PBC must add up to the area of the entire triangle, which is why their proportions must sum to 1.

This definition makes sense to me, and intuitively it seems like the larger $\alpha$ the more A is weighted. I was just wondering why this works mathematically (e.g. why must $\alpha + \beta + \gamma = 1$)?

@julialuo Call the point in question P. The line they drew in the picture separated by (1 - alpha) and alpha is the altitude of the triangle ABC from the vertex A. We can see the length alpha is the altitude of the triangle PBC. Thus, the area of ABC is 1/2 * b * h or 1/2 * BC * 1 and the area of PBC is 1/2 * BC * alpha, the the proportion of area of PBC to ABC or PBC/ABC is alpha.

We can repeat this comparison of altitudes for the altitudes of ABC drawn from B and C, so we get that beta and gamma are the proportions of the area of PAC and PAB to the entire triangle, respectively. Visually, we can easily see that PAC and PAB and PBC must add up to the area of the entire triangle, which is why their proportions must sum to 1.