In Delta ABC,A-D-B,A-E-C and overline{DE} par | vedclass

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Question

(C) According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two s

Vedclass · Accepted Answer

$EC:BD = 4:3$

Vedclass · Answer

(C) According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.Since $\overline{DE} \parallel \overline{BC}$,we have $\frac{AD}{DB} = \frac{AE}{EC}$.Given $AB:AC = 3:4$,let $AB = 3k$ and $AC = 4k$.By the property of similar triangles,$\Delta ADE \sim \Delta ABC$.Therefore,$\frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC}$.From the given ratio,$\frac{AD}{AB} = \frac{AE}{AC} = \frac{3}{4}$.This implies $AD = 3x$ and $AB = 4x$ (where $AB$ is the total length),but here $AB$ is given as $3k$. Thus,$AD = \frac{3}{4} AB$ and $AE = \frac{3}{4} AC$.Then $BD = AB - AD = AB - \frac{3}{4} AB = \frac{1}{4} AB$.Similarly,$EC = AC - AE = AC - \frac{3}{4} AC = \frac{1}{4} AC$.Now,calculating the ratio $\frac{EC}{BD} = \frac{\frac{1}{4} AC}{\frac{1}{4} AB} = \frac{AC}{AB} = \frac{4}{3}$.Thus,$EC:BD = 4:3$.

In $\Delta ABC$,$A-D-B$,$A-E-C$ and $\overline{DE} \parallel \overline{BC}$. If $AB:AC = 3:4$,which of the following holds true?

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