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Ib mathematics question bank
AB is greater than AC, and the bisector of angle B and angle C intersects at point I, which proves that IB is less than IC.

In ▲AIB, IA+IB & gt;; ab blood type

In ▲AIC,ia+IC >; Alternating current

Because of AB> communication,

So: IA+IB & gt;; IA+IC

So: IB> integrated circuit

It can also be seen from the figure that IB is greater than IC, and the topic should be wrong.

Supplement 1: in ▲ABP, PA+PB & gt;; AB,

In ▲ACP,pa+PC >; Alternating current

In ▲BCP, PC+PB & gt;; B.C.

Because of AB & gtAC, PB> personal computers.

Because AB> is BC, PA> personal computer.

So PA+PB & gt;; 2PC> personal computer

2.In ▲ABD,a b+BD & gt; advertisement

In ▲ACD,AC+CD >; advertisement

In ▲BCE, BC+CE & gt;; exist

In ▲BAE,a b+ AE & gt; exist

In ▲BCF,BC+BF & gt; padre

In ▲ACF,AC+AF >; padre

Add up the above six inequalities, 3 (AB+BC+CA) >; 2(AD+BE+CF)

Namely: 2 (ad+be+cf) < 3(AB+BC+CA)