MathGoldMedalist
Sign InRegister

Toolkit 109

Triangle Proportionality Theorem and its Converse

Triangle Proportionality Theorem

Triangle Proportionality Theorem
DE∥BC⟹ADAB=AEAC=DEBCDE\parallel BC\Longrightarrow\frac{AD}{AB}=\frac{AE}{AC}=\frac{DE}{BC}

Equivalently,

ADDB=AEEC\frac{AD}{DB}=\frac{AE}{EC}

Converse of the Triangle Proportionality Theorem

Converse of the Triangle Proportionality Theorem
ADAB=AEAC⟹DE∥BC\frac{AD}{AB}=\frac{AE}{AC}\Longrightarrow DE\parallel BC

Equivalently,

ADDB=AEEC\frac{AD}{DB}=\frac{AE}{EC}

Proof

Proof of Triangle Proportionality Theorem

Since DE∥BCDE\parallel BC,

∠ADE=∠ABC,∠AED=∠ACB\angle ADE=\angle ABC,\qquad \angle AED=\angle ACB

Therefore, by AA similarity,

△ADE∼△ABC\triangle ADE\sim\triangle ABC
ADAB=AEAC=DEBC\frac{AD}{AB}=\frac{AE}{AC}=\frac{DE}{BC}

For the converse,

Proof of Converse of Triangle Proportionality Theorem
ADDB=AEEC\frac{AD}{DB}=\frac{AE}{EC}

By Ratio Manipulation,

ADAD+DB=AEAE+EC\frac{AD}{AD+DB}=\frac{AE}{AE+EC}
ADAB=AEAC\frac{AD}{AB}=\frac{AE}{AC}
∠DAE=∠BAC\angle DAE=\angle BAC

Therefore, by SAS similarity,

△DAE∼△BAC\triangle DAE\sim\triangle BAC
∠ADE=∠ABC\angle ADE=\angle ABC

Therefore,

DE∥BC.□DE\parallel BC.\quad\square
Back to Toolkit