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In Triangle ABC, point D is on BC is such that AD\perp BC and E is the middle point of BC  and b^2+2 a^2=c^2 distance between D and E is?  

Option: 1

b


Option: 2

\frac{a+b+c}{2}


Option: 3

\frac{b+c}{2}


Option: 4

a


Answers (1)

 

 

Projection Formula -

Projection Formula

In the ΔABC, 

Projection of AB on BC is BD

Projection of AC on BC is DC

 

 

Now, 

BD = c cos B and DC = b cos C

and, BC = a = BD + DC

                    = c cos B + b cos C

a = c cos B + b cos C

In a similar way, other projection formula can be derived

  1. a = c cos B + b cos C

  2. b = c cos A + a cos C

  3. c = b cos A + a cos B

-

 

 

Diagram -

 

DE=EC-DC\\ \text{\ \ \ \ \ \ \ \ \ \ }=\frac{a}{2}-b \cos C\\ \text{\ \ \ \ \ \ \ \ \ \ }=\frac{a}{2}-b \frac{a^2+b^2-c^2}{2ab}\\ \text{\ \ \ \ \ \ \ \ \ \ }=\frac{a}{2}-b\frac{a^2-(c^2-b^2)}{2ab}\\\ \text{\ \ \ \ \ \ \ \ \ \ }=\frac{a}{2}-\frac{a^2-(2a^2)}{2a}\\\ \text{\ \ \ \ \ \ \ \ \ \ }=a

Posted by

Ramraj Saini

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