Prove the Following:

Question

Given: BI = IN

Prove: AI > IR

Answer 1

Here is one of many possible contradictions,

which is based on the fact that, in any

triangle, the side opposite the larger

angle MUST be larger.

Suppose AI < IR

then angle ARI < angle IAR

But IAR and PAN are vertical angles,

so IR = PN

Then IA < NP

which forces angle IRA < angle NAP

which forces angle IRA < IAR

Now BI = IN, which forces

isoceles triangle BIN.

Then BR + RI = IA + AN

BR - AN = IA - RI < 0 since IA<IR

then BR< AN

This is a contradiction with respect

to triangles PAN and PRB, of which,

AN and BR are opposite sides of angle

P. The sides should at least be

proportional and increasing, yet

we have BR < AN, a contradiction.

THerefore AI > IR must hold