Question

ABC is a right angle triangle at B.If D and E are mid point of BC and AB respectively prove that AD^2+CE^2=5DE^2

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Answer 1

Step-by-step explanation:

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Answer 2

Answer:

Given:

• ABC is a right angled triangle at B.

• D and E are mid point of BC and AB respectively.

To prove:

• AD² + CE² = 5DE²

Solution:

AB = 2BE = 2AE ... since E is midpoint of seg AB.

BC = 2BD = 2CD ... since D is midpoint of seg BC.

AC = 2ED

Reason:

Line formed by joining midpoints of two side of triangle is always parallel to third side and also half of the third side.

In ∆ ABC,

(i) AC² = AB² + BC².... By Pythagoras theorem

In ∆ ABD,

(ii) AD² = AB² + BD²... By Pythagoras theorem

In ∆ EBC,

(iii) CE² = BE² + BC²... By Pythagoras theorem

In ∆ EBD,

(iv) DE² = EB² + BD²

Add equations (ii) and (iii), we get

=> AD² + CE² = AB² + BD² + BE² + BC²

=> AD² +CE² = AB² + BC² + BD² + BE²

... from (i)

=> AD² + CE² = AC² + BD² + BE²

... from (iv)

=> AD² + CE² = AC² + DE²

Substitute AC = 2DE

=> AD² + CE² = (2DE)² + DE²

=> AD² + CE² = 4DE² + DE²

=> AD² + CE² = 5DE²

Hence, proved.