Question

Basic proportionality Theron

Answer 1

Basic Proportionality theorem

Statement:-

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

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Basic  Proportionality Theorem Proof

☆ Let us now try to prove the basic proportionality theorem statement

☆ Consider a triangle ΔABC, as shown in the given figure. ☆ In this triangle, we draw a line PQ parallel to the side BC of ΔABC and intersecting the sides AB and AC in P and Q, respectively.

☆ According to the basic proportionality theorem as stated above, we need to prove:

$\bf \:  ⟼ \dfrac{AP}{PB} = \dfrac{AQ}{QC}$

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Construction

Join the vertex B of ΔABC to Q and the vertex C to P to form the lines BQ and CP and then drop a perpendicular QN to the side AB and also draw PM⊥AC as shown in the given figure.

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Proof :-

$\sf \:  Now, \: the \: area \: of \: ∆APQ = \dfrac{1}{2} \times AP \times QN$

☆ (Since, area of a triangle= 1/2× base × height)

$\sf \:  Similarly, \: area \: of \: ∆PBQ=  \dfrac{1}{2}  × PB × QN$

$\sf \:  Also, \: area \: of \: ∆APQ =  \: \dfrac{1}{2} × AQ × PM$

$\sf \:  Also, \: area \: of \: ∆QCP = \dfrac{1}{2}  × QC × PM$

☆ Now, if we find the ratio of the area of triangles ∆APQand ∆PBQ, we have

$\sf \:  \dfrac {area ~of~ ∆APQ}{area~ of~ ∆PBQ} = \dfrac {\frac 12 ~×~AP~×~QN}{ \frac 12~×~PB~×~QN}$

$\bf\implies \: \dfrac {area ~of~ ∆APQ}{area~ of~ ∆PBQ} \: = \: \dfrac {AP}{PB} - - - (1)</p><p>$

☆ Now, if we find the ratio of the area of triangles ∆APQand ∆QCP, we have

$\sf \:  \dfrac {area~ of~ ∆APQ}{area~ of~ ∆QCP} \: = \dfrac {\frac12~ ×~ AQ~×~PM}{\frac 12 ~×~QC~×~PM}$

$\bf\implies \: \dfrac {area~ of~ ∆APQ}{area~ of~ ∆QCP} \: = \dfrac {AQ}{QC} - - (2)</p><p>$

☆ According to the property of triangles, the triangles drawn between the same parallel lines and on the same base have equal areas.

☆ Therefore, we can say that ∆PBQ and QCP have the same area.

$\bf\implies \:area \: of \: ∆PBQ = area \: of \: ∆QCP \: - - (3)$

☆ Therefore, from the equations (1), (2) and (3) we can say that,

$\bf \:  ⟼ \dfrac{AP}{PB} = \dfrac{AQ}{QC}$

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$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$