Question

In the figure, seg AD I side BC
and B-D-C then prove that
AB^2 - BD^2 = AC^2 - CD^2.
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Answer 1

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Answer 2

Given: The segment AD is a perpendicular to the side BC of ΔABC, i.e., AD⊥BC.

To prove:  $AB^2-BD^2=AC^2-CD^2$

Step-by-step explanation:

Now, we know that the segment AD is a perpendicular to the side BC of  ΔABC, the triangles, ΔADB and ΔACD, are right angled triangles.

Now, we know that for right angles triangles, the pythagoras theorem states that the square of hypotenuse is the sum of squares of perpendicular and the base of triangle, i.e., $H^2=P^2+B^2$

Applying the pythagoras theorem in ΔADB, we get:

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

or we can say:

$AD^2=AB^2-BD^2$       ...(i)

Similarly, applying the pythagoras theorem in ΔACD, we get:

$Ac^2=AD^2+CD^2$

or we can say:

$AD^2=AC^2-CD^2$       ...(ii)

Equating the equations (i) and (ii), we get:

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

Hence proved.