# In a #DeltaABC#, right angled at #A#, a point #D# is on side #AB#. Prove that #CD^2=BC^2+BD^2#?

With a point

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To prove (CD^2 = BC^2 + BD^2) in triangle (ABC), where (ABC) is a right-angled triangle with right angle at (A), and (D) is a point on side (AB):

Using the Pythagorean theorem, we know that in a right-angled triangle, the square of the length of the hypotenuse ((BC)) is equal to the sum of the squares of the lengths of the other two sides.

So, in triangle (BCD), we have: [BC^2 = BD^2 + CD^2]

Rearranging the terms, we get: [CD^2 = BC^2 - BD^2]

This matches the desired equation (CD^2 = BC^2 + BD^2). Therefore, the statement is proven.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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