1. ## geometry -Euclidean

Consider triangle ABC,suppose D is midpoint of AC and that the line through B perpendicular to AC intersects AC at a point X lying between D and C using Pythagoras theorem prove

$|BA|^{2}+|BC|^{2}=2|BD|^{2}+2|AD|^{2}$

i drew pretty nasty diagram couldn't work out where or wether to use SAS ASA or if any more criterion,axioms involved,thanks.

2. It's tedious, but it's not tricky. Where are you stuck?

It appears to me that your instructsion are to use the Pythagorean Theorem. Why are you worried about congruent triangles in any other sense.

For simplicity, I labled the pieces like this:

AB = a
BC = b
CX = c
XD = d
DA = e
DB = f
BX = g

The, by the Pythagorean Theorem, we have:

(e+d)^2 + g^2 = a^2

d^2 + g^2 = f^2

c^2 + g^2 = b^2

Also given in the problem statement is:

e = c + d

That's all you need to show a^2 + b^2 = 2f^2 + 2e^2

'c' does not appear in the final expression, so get rid of it.

c = e - d

Substituting into the other three equations gives:

(e+d)^2 + g^2 = a^2

d^2 + g^2 = f^2

(e-d)^2 + g^2 = b^2

'g' doesn't appear in the final expression, so get rid of it.

g^2 = f^2 - d^2

Substituting into the other two equations gives:

(e+d)^2 + f^2 - d^2 = a^2

(e-d)^2 + f^2 - d^2 = b^2

'd' doesn't appear in the final expression, so get rid of it. It's not quite as obvious how to do this. Just expand everything and see what happens.

e^2 + 2ed + f^2 = a^2

e^2 - 2ed + f^2 = b^2

Hey!! Where did the d^2 go?! That was SWEET!

Now what? You're not going to make me do ALL the work, are you? You still need to get rid of the 'd'.

Let's see what you get.

3. 4ed=a^2 + b^2

and sub what they are equal to, thanks ever so much,think key is knowing to label e as c+d or whatever.thanks again