Proportion and similarity question

[Jasonbot]

Hi everyone. COuld you guys please help me with this simple geometry problem?

In triangleADC, angleD=90*, DB perpendicular AC, BK||AD and 2DK=KC
Using triangleDAB ||| triangleCAD (already proven using AAA) prove that 2DA=sqrt3.BC

thanks,
Jason

 

[Jasonbot]

Don't they teach geometry in foreign countries?

 

[swiftaw]

OK,

AB^2 + BD^2 = AD^2
BC^2 + BD^2 = CD^2

Putting these together we get that

AB^2 + CD^2 - BC^2 = AD^2 (*)

Now, by similarity we have that AC = 3AB, so AC^2 = 9AB^2.

Using that fact (*) becomes
(1/9)AC^2 + CD^2 - BC^2 = AD^2

But AC^2 = AD^2 + CD^2, thus

(1/9)(AD^2 + CD^2) + CD^2 - BC^2 = AD^2

Re-arranging we get
(10/9)CD^2 - BC^2 = (8/9)AD^2

Now, use the fact that CD^2 = AC^2 - AD^2 we get

(10/9)(AC^2 - AD^2) - BC^2 = (8/9)AD^2
or, in other words
2AD^2 = (10/9)AC^2 - BC^2 (**)

But, AC = (3/2)BC, so substituting into (**) gets us

2AD^2 = (10/9)(9/4)BC^2 - BC^2 = (6/4)BC^2

Doubling both sides

4AD^2 = (12/4)BC^2 = 3BC^2

Taking square roots

2AD = sqrt(3)*BC

 

[Jasonbot]

Thank you so much! I have no idea how you see these things but I'm sure Ill get it DO you look at what you're supposed to prove and work backwards?

 

[swiftaw]

Thank you so much! I have no idea how you see these things but I'm sure Ill get it DO you look at what you're supposed to prove and work backwards?

No, that is bad form. The only thing I looked in the final answer was that it only contained AD and BC. I started by writing down things I knew to be true and then tried to combine them in such a way to leave only AD and BC.