Answer
Verified
410.1k+ views
Hint: First, we draw a triangle ABC and draw straight line DE passes through two sides of the triangle and parallel to the third side. Now, we have to prove that the ratio of AD and DB is equal as it is divided by point D. By using this methodology, we can easily solve our problem.
Complete step-by-step answer:
Statement: If a line passes through two sides of a triangle, then it is parallel to the third side that divides the other two sides in the same ratio.
First, we draw a triangle ABC and draw straight line DE passes two sides of the triangle and parallel to the third side. This can be shown as:
In $\Delta ADE$, considering base to be AD and height to be ME.
Area of triangle ADE$=\dfrac{1}{2}\times AD\times ME...(1)$
Again, taking triangle ADE, but this time base is AE and height is ND.
In $\Delta ADE$,
Area of triangle ADE$=\dfrac{1}{2}\times AE \times ND...(2)$
In $\Delta BDE$, where base is DB and height is ME:
Area of triangle BDE$=\dfrac{1}{2}\times DB\times ME...(3)$
In $\Delta DEC,$ where base is EC and height is ND
Area of triangle DEC$=\dfrac{1}{2}\times EC\times ND...(4)$
Now, we divide Equation (1) and (3) to obtain equation (5)
$\begin{align}
& \dfrac{area(\Delta ADE)}{area(\Delta BDE)}=\dfrac{\dfrac{1}{2}\times AD\times ME}{\dfrac{1}{2}\times DB\times ME} \\
& \dfrac{area(\Delta ADE)}{area(\Delta BDE)}=\dfrac{AD}{DB}...(5) \\
\end{align}$
Now, divide equation (2) and equation (4) to obtain equation (6)
$\begin{align}
& \dfrac{area(\Delta ADE)}{area(\Delta DEC)}=\dfrac{\dfrac{1}{2}\times AE\times ND}{\dfrac{1}{2}\times EC\times ND} \\
& \dfrac{area(\Delta ADE)}{area(\Delta BDE)}=\dfrac{AE}{EC}...(6) \\
\end{align}$
By using the theorem,
$\begin{align}
& area\left( \Delta ADE \right)=area\left( \Delta DEC \right) \\
& \dfrac{area(\Delta ADE)}{area(\Delta BDC)}=\dfrac{area(\Delta ADE)}{area(\Delta DEC)}=\dfrac{AD}{BD} \\
& \dfrac{area(\Delta ADE)}{area(\Delta DEC)}=\dfrac{AD}{BD}...(7) \\
\end{align}$
By the equation (6) and equation (7), we can rewrite it as
\[\begin{align}
& \dfrac{AE}{EC}=\dfrac{AD}{BD} \\
& \therefore \dfrac{AD}{BD}=\dfrac{AE}{EC} \\
\end{align}\]
Hence, we proved the existence of BPT.
Note: The key concept involved in solving this problem is the knowledge of the area of the triangle by considering different bases and height corresponding to that base. By using this consideration, we manipulate the obtained equations to prove the existence of BPT.
Complete step-by-step answer:
Statement: If a line passes through two sides of a triangle, then it is parallel to the third side that divides the other two sides in the same ratio.
First, we draw a triangle ABC and draw straight line DE passes two sides of the triangle and parallel to the third side. This can be shown as:
In $\Delta ADE$, considering base to be AD and height to be ME.
Area of triangle ADE$=\dfrac{1}{2}\times AD\times ME...(1)$
Again, taking triangle ADE, but this time base is AE and height is ND.
In $\Delta ADE$,
Area of triangle ADE$=\dfrac{1}{2}\times AE \times ND...(2)$
In $\Delta BDE$, where base is DB and height is ME:
Area of triangle BDE$=\dfrac{1}{2}\times DB\times ME...(3)$
In $\Delta DEC,$ where base is EC and height is ND
Area of triangle DEC$=\dfrac{1}{2}\times EC\times ND...(4)$
Now, we divide Equation (1) and (3) to obtain equation (5)
$\begin{align}
& \dfrac{area(\Delta ADE)}{area(\Delta BDE)}=\dfrac{\dfrac{1}{2}\times AD\times ME}{\dfrac{1}{2}\times DB\times ME} \\
& \dfrac{area(\Delta ADE)}{area(\Delta BDE)}=\dfrac{AD}{DB}...(5) \\
\end{align}$
Now, divide equation (2) and equation (4) to obtain equation (6)
$\begin{align}
& \dfrac{area(\Delta ADE)}{area(\Delta DEC)}=\dfrac{\dfrac{1}{2}\times AE\times ND}{\dfrac{1}{2}\times EC\times ND} \\
& \dfrac{area(\Delta ADE)}{area(\Delta BDE)}=\dfrac{AE}{EC}...(6) \\
\end{align}$
By using the theorem,
$\begin{align}
& area\left( \Delta ADE \right)=area\left( \Delta DEC \right) \\
& \dfrac{area(\Delta ADE)}{area(\Delta BDC)}=\dfrac{area(\Delta ADE)}{area(\Delta DEC)}=\dfrac{AD}{BD} \\
& \dfrac{area(\Delta ADE)}{area(\Delta DEC)}=\dfrac{AD}{BD}...(7) \\
\end{align}$
By the equation (6) and equation (7), we can rewrite it as
\[\begin{align}
& \dfrac{AE}{EC}=\dfrac{AD}{BD} \\
& \therefore \dfrac{AD}{BD}=\dfrac{AE}{EC} \\
\end{align}\]
Hence, we proved the existence of BPT.
Note: The key concept involved in solving this problem is the knowledge of the area of the triangle by considering different bases and height corresponding to that base. By using this consideration, we manipulate the obtained equations to prove the existence of BPT.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Define absolute refractive index of a medium
Find out what do the algal bloom and redtides sign class 10 biology CBSE
Prove that the function fleft x right xn is continuous class 12 maths CBSE
Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Which type of bond is stronger ionic or covalent class 12 chemistry CBSE
What organs are located on the left side of your body class 11 biology CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
How fast is 60 miles per hour in kilometres per ho class 10 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
When people say No pun intended what does that mea class 8 english CBSE