Question

State and prove the BPT theorem.

Answer 1

$\huge{\textsf{\textbf{\color{lime}{Statement:-}}}}$

Basic Proportionality Theorem states that, if a line is parallel to a side of a triangle which intersects the other sides in two distinct points, then the line divides those sides of the triangle in proportion.

$\huge{\textsf{\textbf{\color{cyan}{Given:-}}}}$

Let ABC be the triangle.

The line l parallel to BC intersect AB at D and AC at E.

$\huge{\textsf{\textbf{\color{lime}{To\:prove:-}}}}$

We have to prove that :

$\frac{AD}{DB} = \frac{AE}{EC}$

$\huge{\textsf{\textbf{\color{cyan}{Construction:-}}}}$

Join BE, CD.

Draw EF perpendicular to AB, DG perpendicular to CA.

$\huge{\textsf{\textbf{\color{lime}{Proof:-}}}}$

Since, EF is perpendicular to AB,

EF is the height of triangles ADE and DBE.

$Area \: of \: a \: triangle =\: \frac{1}{2} \times base \times height \\Area \: of \: ∆ADE = = > \frac{1}{2} \times AD \times EF \\Area \: of \: ∆DBE= \frac{1}{2} \times DB \times EF$

$\frac{ Area \: of \: ∆ADE }{Area \: of \: ∆DBE} = \frac{ \frac{1}{2} \times AD \times EF}{ \frac{1}{2} \times DB \times EF } \\ = = > \frac{AD}{DB} ....(1)$

Similarly,

$\frac{Area \: of \: ∆ADE}{Area \: of \: ∆DCE} = \frac{ \frac{1}{2}× AE×DG}{ \frac{1}{2} \times EC×DG } \\ = = > \frac{AE}{EC} ....(2)$

But ∆DBE and ∆DCE are on the same base DE and between the same parallel straight line BC and DE.

Area of ∆DBE = Area of ∆DCE....(3)

From (1), (2) and (3) we have :

$\large{\frac{AD}{DB} = \frac{AE}{EC}}$

Hence proved !!

Answer 2

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Step-by-step explanation:

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