Triangle in circle

[psi20]

How do you prove that AD * AC = AB^2?

[psi20]

AD is the diameter of the circle, ABC is a right triangle, etc. etc.

[Primarygun]

Neither A nor B is equal to 90 degree, so C must be 90 degree.

Prove BC^2=ACxCD first, then you can prove the orginal one.

There are two methods.

(1)Extend BC to E where E is a point on the circumference.

Since angle C=90 degree, then it must passes through the centre of the circle.

So BC=CE, then you can prove it with Pyth. theorem.

You may consider the two similar triangle, the ones like a butterfly.

(2)Join BD, angle ABD=90 degree.

You can see that ACxCD=BC^2.

 

Either one of the two above leads you the answer.

[Primarygun]

By the way, I thought applied maths is the mathematics applying maths into physics,.....

So I seldom browse it. I like pure maths.

[psi20]

I didn't get that at all.

[DQW]

What part didn't you get ?

 

Let's look at PG's solution #2 again :

 

First join BD (at least mentally).

Since AD is a diameter, <ABD is a right angle ("angle in a semicircle is a right angle")

Hence triangles ACB and BCD are similar (all angles are equal)

So, AC/BC = BC/CD (corresponding sides are in the same ratio)

Thus, AC*CD = BC^2

But from Pythagoras, BC^2 = AB^2 - AC^2 ...

 

...plug in and complete...

[Primarygun]

Thanks to DQW, my prove is completed .

When we prove, we usually do from the backward first.

With the given equation, try to split it into some triangles, some other laws.

For example, if I splitted and found that I need to prove is XY^2+XZ^2=XY^2,

then construct the responsible triangle and then prove angle xyz is equal to 90 degree.