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Proving BE:EX 3:1 in Triangle ABC with Medians
Proving BE:EX 3:1 in Triangle ABC with Medians
This article explores a geometric proof in triangle ABC, where AD is a median on BC, and BE is a median on AD. If BE is further extended to meet AC at point X, the problem is to prove that BE:EX 3:1.
Step-by-Step Proof
Step 1: Drawing Parallel Lines
First, we draw a line XP parallel to BD. By doing so, two similar triangles are formed which will help us in finding the ratios of the segments.
Step 2: Applying Similarity Theorems
This process involves the following steps:
Triangle BDE is similar to Triangle XPE. This gives us the ratio:
$frac{BD}{PX} frac{DE}{PE}$
Triangle ADC is similar to Triangle APX. This gives us the ratio:
$frac{DC}{PX} frac{AD}{AP}$
Step 3: Using the Given Information
Since BD DC, we have:
$frac{AD}{AP} frac{DE}{PE}$
Multiplying both sides by 2 (as AD 2DE):
$frac{2 times DE}{AP} frac{DE}{PE}$
Simplifying this, we get:
$frac{AP}{PE} 2$
So, simplifying further:
$frac{AP}{PE} - 1 2 - 1$
Step 4: Extending BE to Meet AC at X
We extend BE to meet AC at point X. We draw a line DY parallel to BX. Then, we use the following steps:
In Triangle BEC, we have:
$frac{BD}{DC} frac{XY}{YC}$
From the Basic Proportionality Theorem (BPT),
$BD DC$, hence $frac{XY}{YC} 1$
So, $XY YC$ (Equation 1).
In Triangle ADC, we have:
$AX XY$ (Equation 2).
Step 5: Using Equations to Find AX XY YC
From Equations 1 and 2:
$AX XY YC$ (Equation 5).
Step 6: Proving Similarity of Triangles
We prove the similarity of triangles DYC and BXC by the Angle-Angle (AA) rule:
$frac{DY}{BX} frac{DC}{BC}$
Since $DC frac{1}{2}BC$,
$frac{DY}{BX} frac{1}{2}$
So, $2DY BX$ (Equation 3).
Step 7: Proving Similarity of Triangles AEX and ADY
We prove the similarity of triangles AEX and ADY by the AA rule:
$frac{XE}{DY} frac{AX}{AY}$
From Equation 5, $AX 2AX$;
$frac{XE}{DY} frac{1}{2}$
Therefore,
$2XE DY$ (Equation 4).
Step 8: Concluding the Proof
Combining Equations 3 and 4:
$2 times 2XE BX$
$4XE BX$
So,
$BE 3XE$
Hence,
$frac{BE}{EX} 3 : 1$
Hence Proved
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