Question

In the proof of the Basic Proportionality theorem for \(\triangle ABC\) with \(DE \parallel BC\) (on AB, E on AC), the construction made is to join ________.

• A. AD and AE
• B. BD and CE
• C. BE and CD
• D. AB and DE

Hint:

Visualize the triangle with the parallel line \(DE\).
The vertices to cross-connect are the lower vertices of the smaller upper triangle (\(D\) and \(E\)) with the lower vertices of the main triangle (\(C\) and \(B\)).
This forms a 'cross' shape in the lower trapezoid part: \(BE\) and \(CD\).

Answer 1

The Correct Option is C

Step 1: Think about what the proof is trying to compare.
The Basic Proportionality Theorem proof compares the ratio $\frac{AD}{DB}$ with the ratio $\frac{AE}{EC}$ by looking at areas of triangles built on segment $DE$.
Step 2: Work out which triangles are needed.
To compare $AD$ with $DB$ using areas, we need a triangle on base $DB$ with the same height as $\triangle ADE$, which means connecting $E$ to $B$. Similarly, to compare $AE$ with $EC$, we need a triangle on base $EC$ with the same height as $\triangle ADE$, meaning connecting $D$ to $C$.
Step 3: See why these specific joins matter.
Joining $BE$ gives triangle $BDE$ sitting on base $DE$, and joining $CD$ gives triangle $CDE$, also on base $DE$, between the same parallels $DE$ and $BC$. This shared base and shared parallels is exactly what makes their areas equal in the next step of the proof.
Step 4: State the construction.
So the segments drawn are BE and CD.
\[ \boxed{\text{BE and CD}} \]