Difference between revisions of "Ceva I.2"
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− | We want to prove <math>X,Y,B</math> [[collinear]], so we consider from which which direction we want to prove this. We can prove <math>\angle XYC + \angle BYC = 180</math> to do it, but we don't know any angles, so that probably won't be much use. Instead, we can prove <math>CM, DP, XB</math> collinear, since the intersection of <math>CM</math> and <math>DP</math> is <math>Y</math>. So, let's consider [[Ceva's]] (a concurrency related formula) on <math>\triangle BCP</math>. | + | We want to prove <math>X,Y,B</math> [[collinear]], so we consider from which which direction we want to prove this. We can prove <math>\angle XYC + \angle BYC = 180</math> to do it, but we don't know any angles, so that probably won't be much use. Instead, we can prove <math>CM, DP, XB</math> collinear, since the intersection of <math>CM</math> and <math>DP</math> is <math>Y</math>. So, let's consider [[Ceva's Theorem]] (a concurrency related formula) on <math>\triangle BCP</math>. |
Let <math>AB = c, AC = b, BC = a</math>. That means <math>AM = MB = \frac{c}{2}</math>. There are a lot of unknowns here, so let further set <math>AP = x, CD = y</math>. We know that | Let <math>AB = c, AC = b, BC = a</math>. That means <math>AM = MB = \frac{c}{2}</math>. There are a lot of unknowns here, so let further set <math>AP = x, CD = y</math>. We know that | ||
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Now, if we extend <math>EM</math> through <math>E</math> and intersect the line at <math>C</math> [[parallel]] to <math>AB</math> at point <math>G</math>, we see <math>\triangle GEC \sim \triangle MEA</math>. Thus, <math>\dfrac{GC}{AM} = \dfrac{EC}{AE} = \dfrac{CD}{BD} \implies \dfrac{GC}{\frac{c}{2}} = \dfrac{y}{a-y} \implies GC = \dfrac{\frac{c}{2} \cdot y}{a-y}</math>. Using <math>\triangle GCX \sim \triangle MPX</math>, <math>\dfrac{CX}{XP} = \dfrac{GC}{PM} = \dfrac{\frac{\frac{c}{2} \cdot y}{a-y}}{\frac{c}{2}-x}</math>. Thus, | Now, if we extend <math>EM</math> through <math>E</math> and intersect the line at <math>C</math> [[parallel]] to <math>AB</math> at point <math>G</math>, we see <math>\triangle GEC \sim \triangle MEA</math>. Thus, <math>\dfrac{GC}{AM} = \dfrac{EC}{AE} = \dfrac{CD}{BD} \implies \dfrac{GC}{\frac{c}{2}} = \dfrac{y}{a-y} \implies GC = \dfrac{\frac{c}{2} \cdot y}{a-y}</math>. Using <math>\triangle GCX \sim \triangle MPX</math>, <math>\dfrac{CX}{XP} = \dfrac{GC}{PM} = \dfrac{\frac{\frac{c}{2} \cdot y}{a-y}}{\frac{c}{2}-x}</math>. Thus, | ||
<cmath>\dfrac{PM}{MB} \cdot \dfrac{BD}{DC} \cdot \dfrac{CX}{XP} = \dfrac{\frac{c}{2}-x}{\frac{c}{2}} \cdot \dfrac{a-y}{y} \cdot \dfrac{CX}{XP} = \dfrac{\frac{c}{2}-x}{\frac{c}{2}} \cdot \dfrac{a-y}{y} \cdot \dfrac{\frac{\frac{c}{2} \cdot y}{a-y}}{\frac{c}{2}-x}=1</cmath> | <cmath>\dfrac{PM}{MB} \cdot \dfrac{BD}{DC} \cdot \dfrac{CX}{XP} = \dfrac{\frac{c}{2}-x}{\frac{c}{2}} \cdot \dfrac{a-y}{y} \cdot \dfrac{CX}{XP} = \dfrac{\frac{c}{2}-x}{\frac{c}{2}} \cdot \dfrac{a-y}{y} \cdot \dfrac{\frac{\frac{c}{2} \cdot y}{a-y}}{\frac{c}{2}-x}=1</cmath> | ||
+ | |||
+ | <math>\mathcal{QED}</math> |
Latest revision as of 16:38, 26 February 2023
Problem
- Let
be the midpoint of side
of triangle
. Points
and
lie on line segments
and
, respectively, such that
and
are parallel. Point
lies on line segment
. Lines
and
intersect at
and lines
and
meet at
. Prove that
are collinear.
Solution
We want to prove collinear, so we consider from which which direction we want to prove this. We can prove
to do it, but we don't know any angles, so that probably won't be much use. Instead, we can prove
collinear, since the intersection of
and
is
. So, let's consider Ceva's Theorem (a concurrency related formula) on
.
Let . That means
. There are a lot of unknowns here, so let further set
. We know that
Now, if we extend
through
and intersect the line at
parallel to
at point
, we see
. Thus,
. Using
,
. Thus,