Question

a line parallel to one side of a triangle divides the other two sides in equal propotion​

Answer 1

Theorem :

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points then prove that the two sides are divided in the same ratio.

Given:

DE // BC

To prove that :

EC/AE = BD/AD

Proof:

∠AED=∠ACB Corresponding angles

∠ADE=∠ABC Corresponding angles

∠EAD is common to both the triangles

⇒ΔAED∼ΔACB by AAA similarity

⇒ AC/AE = AB/AD

⇒ AE+EC/AE = AD+BD/AD

⇒EC/AE = BD/AD

Hence proved

Answer 2

Answer:

Basic proportionality theorem

Step-by-step explanation:

If a line parallel to a side of a triangle intersects the remaining sides in two distinct points, then the line divides the sides in same propotion.

Given : In /\ ABC, line PQ || side BC, line PQ intersects sides AB & AC in points P & Q respectively.

To prove : AP/PB = AQ/QC.

Construction : Draw seg PC & seg BQ.

Proof : In /\ APQ & /\ BPQ,

A(/\ APQ) / A(/\ BPQ) = AP/PB_ _ _ (1)_ _ _ [Triangles having equal heights.]

Then,

In /\ APQ & /\ CPQ,

A(/\ APQ) /A(/\ CPQ) = AQ/QC_ _ _ (2)_ _ _ [Triangles having equal heights.]

In /\ BPQ & /\ CPQ,_ _ _ (PQ || BC)_ _ _ (Given)

:. A(/\ BPQ) = A(/\ CPQ) _ _ _ (3)_ _ _ [Triangles having equal height and common base PQ]

Now,

A(/\ APQ) / A( /\ BPQ) = A( /\ APQ) / A( /\ CPQ) _ _ _ [From (1), (2) & (3)]

Therefore,

AP/PB = AQ/QC. _ _ _ _ _ [From (1) & (2)]

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