Question

in the given figure if ad is perpendicular to BC prove that a b square + C D square equal to BD square + AC square

Answer 1

here is your answer OK


In the image above:

ABD is a right angled traingle
and AB is the hypotaneous

(AB)^2 = (AD)^2 + (BD)^2

(AB)^2 - (BD)^2 = (AD)^2 ---------(i)


now,
ACD is also a right angled traingle
(AC)^2 = (AD)^2 + (DC)^2
(AC)^2 - (DC)^2 = (AD)^2 ----------(ii)

(i) = (ii)

(AB)^2 - (BD)^2 =(AC)^2 - (DC)^2
(AB)^2 +(DC)^2 = (AC)^2 + (BD)^2

Answer 2

Hi!

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Given :

AD ⊥ BC

To Prove :

$AB^2 + CD^2 = BD^2 + AC^2$

Proof :

From ΔADC

$AC^2 = AD^2 + CD^2$ ----> 1

From ΔADB

$AB^2 = AD^2 + BD^2$ -----> 2

Subtracting (1) from (2) , We have

$AB^2 - AC^2 = BD^2 - CD^2$

$AB^2 + CD^2 = BD^2 + AC^2$

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Thanks !!