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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.
Construction
Join the vertex B of {tex}\triangle{/tex}ABC 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{tex}\perp{/tex}AC as shown in the given figure.

Proof
Now the area of {tex}\triangle{/tex}APQ = {tex}\frac {1}{2}{/tex} {tex}\times{/tex} AP {tex}\times{/tex} QN (Since, area of a triangle = {tex}\frac {1}{2}{/tex} {tex}\times{/tex} Base {tex}\times{/tex} Height)
Similarly, area of {tex}\triangle{/tex}PBQ = {tex}\frac {1}{2}{/tex} {tex}\times{/tex} PB {tex}\times{/tex} QN
area of {tex}\triangle{/tex}APQ = {tex}\frac {1}{2}{/tex} {tex}\times{/tex} AQ {tex}\times{/tex} PM
Also, area of {tex}\triangle{/tex}QCP = {tex}\frac {1}{2}{/tex} {tex}\times{/tex} QC {tex}\times{/tex} PM ...(i)
Now, if we find the ratio of the area of triangles {tex}\triangle{/tex}APQand {tex}\triangle{/tex}PBQ, we have
{tex}\frac{\text { area of } \Delta A P Q}{\text { area of } \Delta P B Q}{/tex} = {tex}\frac{\frac{1}{2} \times A P \times Q N}{\frac{1}{2} \times P B \times Q N}=\frac{A P}{P B}{/tex}
Similarly, {tex}\frac{\text { area of } \Delta A P Q}{\text { area of } \Delta Q C P}{/tex} = {tex}\frac{\frac{1}{2} \times A Q \times P M}{\frac{1}{2} \times Q C \times P M}=\frac{A Q}{Q C}{/tex} ...(ii)
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 {tex}\triangle{/tex}PBQ and {tex}\triangle{/tex}QCP have the same area.
area of {tex}\triangle{/tex}PBQ = area of {tex}\triangle{/tex}QCP ...(iii)
Therefore, from the equations (i), (ii) and (iii) we can say that,
{tex}\frac{{AP}}{{PB}} = \frac{{AQ}}{{QC}}{/tex}
Also, {tex}\triangle{/tex}ABC and {tex}\triangle{/tex}APQ fulfil the conditions for similar triangles, as stated above. Thus, we can say that {tex}\triangle{/tex}ABC {tex} \sim {/tex} {tex}\triangle{/tex}APQ.
The MidPoint theorem is a special case of the basic proportionality theorem.
According to mid-point theorem, a line drawn joining the midpoints of the two sides of a triangle is parallel to the third side.
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105 C, Chatterjee Lane,
Kolkata
Date: 06.11.2020
To
The Hostel Warden
Shakti Mandir Public School,
Darjeeling
Sub: Requesting to take urgent action on the issue of bullying
Sir,
I am Rajni Roy, sister of Paresh Biswas who is a student of Shakti Mandir Public School.
I have received a letter from my brother who is staying in your hostel. He conveyed that he is being bullied by some senior students who threatened him with dire consequences if he complained. I request you to take proper action on it.
I would be obliged if you consider the situation and take a step against this.
Yours faithfully,
Ranjani
Sia ? 4 years, 8 months ago
C 2/8, Ankur Enclave
New Delhi.
15 August 2020
The Municipal Commissioner
Municipal Office
New Delhi
Subject: Strict action against the insanitary conditions in the colony which can cause diseases.
Respected Madam/Sir
I am a resident of the the address given above. I would like to draw your kind attention to the unsanitary conditions in my colony, the trespassers just throw garbage here and there because there is no dustbin. The water of the drain is coming to the roads and giving it an unpleasant smell , mosquitoes are breeding in the water and flies roam around the garbage, which is also very unhygienic.
When we complained to head of the cleaning department , they said that the cleaners haven't been coming to work for many days and neither do they do their work sincerely.
In these conditions there is a chance of getting multiple diseases which also threatens the life of the residents. I request you to take some strict actions against it.
We would be grateful, if you could help.
Yours sincerely
Deepak/Deepa
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