Question

Triangle ABC is right angled at B and side AB is divided into three equal parts by point D and E, find AC^2-EC^2/DC^2-BC^2​

Answer 1

Answer:

2

Step-by-step explanation:

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Answer 2

Triangle

Given:

ABC is right angled triangle, with right angle at B.

AB is divided into 3 equal parts.

Point D and E are on line AB.

To find:

$\frac{AC^2-EC^2}{DC^2-BC^2}$

Explanation:

D & E are points of trisection of the side AB.

Let $BE=DE=AD=k$

$BD=2k\\\\AB=3k$

We can write

$AC^{2}=(3k)^{2}+ BC^{2} \\\\EC^{2}=k^{2}+ BC^{2} \\\\DC^{2}=(2k)^{2}+ BC^{2}$

So, on putting these values in the

$\frac{AC^2-EC^2}{DC^2-BC^2}$

$= \frac{((3k)^{2}+ BC^{2} )-(k^{2}+BC^{2})}{((2k)^{2}+BC^{2})-BC^{2} }$

$=\frac{9k^2 +BC^{2}-k^2-BC^{2} }{4k^{2} +BC^{2}-BC^{2} }$

$=\frac{8k^2}{4k^2}\\\\=2$

Hence the required value is 2.