Three cevians divided the triangle into six triangles, the areas of which are marked in the figure. 1) Prove that $S_1 \cdot S_2 \cdot S_3 =Q_1 \cdot Q_2 \cdot Q_3$. 2) Determine whether it is true that if $S_1 = S_2 = S_3$, then $Q_1 = Q_2 = Q_3$.
Problem
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Tags: geometry, areas
anduran
28.02.2024 19:19
We want to prove $$\frac{Q_1}{S_1}\cdot \frac{Q_2}{S_2} \cdot \frac{Q_3}{S_3}=1$$Because the pairwise $Q_i, S_i$ have same height, it’s a ratio of their bases, which is true by Ceva’s theorem.
AbhayAttarde01
29.02.2024 02:07
if the lines are cevians, then that means that
$Q_1= S_2$,
$Q_2 = S_3$,
and $Q_3 = S_1$.
by multiplying on both sides, we get
$\boxed{S_1 * S_2 * S_3 = Q_1 * Q_2 * Q_3}$,
which is what we wanted to prove.
(this is probably the wrong way to go about this)
eg4334
29.02.2024 02:43
AbhayAttarde01 wrote:
if the lines are cevians, then that means that
$Q_1= S_2$,
$Q_2 = S_3$,
and $Q_3 = S_1$.
by multiplying on both sides, we get
$\boxed{S_1 * S_2 * S_3 = Q_1 * Q_2 * Q_3}$,
which is what we wanted to prove.
(this is probably the wrong way to go about this)
Cevians \neq medians @2above has good proof though
