Geometry

[Sirotnikov]

So its a basic and silly geometry question, but its late at night and i need assistance.

I have this triangle which is Isosceles (has 2 identicle sides)
It is given that AB = AC.


D is a random point on the base BC.


I need to prove that for every D, AC will be greater than AD.

[Sirotnikov]

So I go like this.

AC = AB.

now I want to prove that AB is larger than AD.

Easy peasy...

We show that D₁ (the right D angle) is the largest angle in the triangle, this it has to be against the largest side, thus AB is necissarily larger than AD


So I prove that angle D₁ is larger than angle B:

D₁ = A₂ + C
given that C = B we get
D₁ = A₂ + B
therefore D₁ is necessarily larger than B.


But - how do I prove that D₁ is larger than A₁???

[Az]

ugh, 2 am, head hurts. is trig allowed?

[Sirotnikov]

hey dude
i miss u


no, not so far. no trig.

i'm probably overseeing some stupid fact.

i'm thinking about drawing an aid to either D` on the other side, or E where E would be the middle of BC.
But it doesn't do good.

[VetLegion]

Originally posted by Sirotnikov
hey dude
i miss u

Hey I was thinking of trying to reply but if you two are having a private conversation...

[civman2000]

EDIT: Read the next post; Mercator says the same thing more succinctly and legibly.

Here's a much simpler proof:
Let E be the midpoint of BC; since the triangle is isosceles, AEB is a right angle. Then AD^2 = AE^2 + DE^2, and AB^2 = AE^2 + BE^2. But clearly BE > DE, so AB > AD.

[Mercator]

How about this...

AC=AB

Now, draw a line straight down from A, giving you point E.

AB²=AE² + EB²
AD²=AE² + ED²

ED < EB thus, AD < AC/AB

[Az]

edit - I think pythagoras and these guys solved it.

[Sirotnikov]

Originally posted by Mercator
How about this...

AC=AB

Now, draw a line straight down from A, giving you point E.

AB²=AE² + EB²
AD²=AE² + ED²

ED < EC thus, AD < AC/AB

i like civman's proof better...

[Sirotnikov]

Originally posted by Az

But - how do I prove that D1 is larger than A1???

It's not. not necessarily. after, D1 isn't necessarily >90 degrees.

It is, since I always choose D1 as the larger angle, not the right angle...

[Az]

btw, was pythagoras allowed in those lessons?

[Mercator]

Bah! Slower than civman

[Az]

It is, since I always choose D1 as the larger angle, not the right angle...

yeah, I figured that much out as I posted...

[civman2000]

Originally posted by Mercator
Bah! Slower than civman

Yeah, but I like your presentation of it better than mine.

[Sirotnikov]

You DanS'ed me