In triangle $ABC$, $\angle B = 45^\circ$ and $\angle C = 75^\circ$. $D$ is the intersection of the angle bisector of $\angle A$ and side $BC$. We need to find the ratios $AC:BC$, $AD:BD$, $BD:BC$, $AB:AC$, $\angle AO_1 D$, $\angle AO_2 O_1$, $AC:AO_1$, $AO_2:AO_1$, and $\triangle ABC: \triangle AO_1 O_2$, where $O_1$ is the circumcenter of $\triangle ABD$ and $O_2$ is the circumcenter of $\triangle ADC$.

1. Problem Description

In triangle $ABC$, $\angle B = 45^\circ$ and $\angle C = 75^\circ$. $D$ is the intersection of the angle bisector of $\angle A$ and side $BC$. We need to find the ratios $AC:BC$, $AD:BD$, $BD:BC$, $AB:AC$, $\angle AO_1 D$, $\angle AO_2 O_1$, $AC:AO_1$, $AO_2:AO_1$, and $\triangle ABC: \triangle AO_1 O_2$, where $O_1$ is the circumcenter of $\triangle ABD$ and $O_2$ is the circumcenter of $\triangle ADC$.

2. Solution Steps

(1)

Using the Law of Sines in $\triangle ABC$:

$\frac{AC}{\sin B} = \frac{BC}{\sin A}$

$\angle A = 180^\circ - 45^\circ - 75^\circ = 60^\circ$

$AC = \frac{\sin B}{\sin A} BC = \frac{\sin 45^\circ}{\sin 60^\circ} BC = \frac{\frac{1}{\sqrt{2}}}{\frac{\sqrt{3}}{2}} BC = \frac{2}{\sqrt{6}} BC = \frac{\sqrt{6}}{3} BC = \sqrt{\frac{6}{9}} BC = \sqrt{\frac{2}{3}} BC$

So $A = 2$ and $B = 3$.

Using the Law of Sines in $\triangle ABD$:

$\frac{AD}{\sin B} = \frac{BD}{\sin (\angle BAD)}$

$\frac{AD}{\sin 45^\circ} = \frac{BD}{\sin 30^\circ}$

$AD = \frac{\sin 45^\circ}{\sin 30^\circ} BD = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}} BD = \frac{2}{\sqrt{2}} BD = \sqrt{2} BD$

So $C = 2$.

Since $\angle ADC = 180^\circ - \angle DAC - \angle C = 180^\circ - 30^\circ - 75^\circ = 75^\circ$, $\triangle ADC$ is an isosceles triangle, which means $AD = AC$. Thus, $\angle ADC = 75^\circ$, so $D = 7$ and $E = 5$.

Since $AD = AC$, we have $AC = \sqrt{2} BD$.

From $AC = \sqrt{\frac{2}{3}} BC$, $\sqrt{2} BD = \sqrt{\frac{2}{3}} BC$, so $BD = \sqrt{\frac{1}{3}} BC = \frac{1}{\sqrt{3}} BC$.

$BD:BC = 1:\sqrt{3}$.

So $F = 1$ and $G = 3$.

$\frac{AB}{\sin C} = \frac{AC}{\sin B}$

$AB = \frac{\sin C}{\sin B} AC = \frac{\sin 75^\circ}{\sin 45^\circ} AC = \frac{\sin(45^\circ + 30^\circ)}{\sin 45^\circ} AC = \frac{\sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ}{\sin 45^\circ} AC = (\cos 30^\circ + \sin 30^\circ) AC = (\frac{\sqrt{3}}{2} + \frac{1}{2}) AC = \frac{\sqrt{3} + 1}{2} AC$

$AB:AC = \frac{1+\sqrt{3}}{2}:1 = 1: \frac{2}{1+\sqrt{3}} = 1 : \frac{2(1-\sqrt{3})}{(1+\sqrt{3})(1-\sqrt{3})} = 1: \frac{2(1-\sqrt{3})}{1-3} = 1: \frac{2(1-\sqrt{3})}{-2} = 1: (-1+\sqrt{3})$

$AB:AC = 1: (\sqrt{3}-1)$.

So $H = 1$, $I = 3$, $J = 1$.

(2)

Since $O_1$ is the circumcenter of $\triangle ABD$, $\angle AO_1 D = 2\angle B = 2(45^\circ) = 90^\circ$. So $K = 9$ and $L = 0$.

Since $O_2$ is the circumcenter of $\triangle ADC$, $\angle ADC = 75^\circ$, so $\angle AO_2 C = 2(180^\circ - 75^\circ) = 210^\circ$, and $\angle AO_2 D = 2 \angle C = 2(75^\circ) = 150^\circ$, since $A, O_2, D, C$ lie on the same circle.

$\angle O_1 A D = 90^\circ$, and $\angle O_2 A D = 15^\circ$.

Since $\angle O_1 A D = 90$, then $AO_1 = AD / \cos(\angle O_1 A D)$.

Consider $\triangle AO_1 O_2$. $\angle AO_2 O_1 = 30^\circ$

$AC = \sqrt{2}AO_1$ so $O=2$.

$AO_2 = \sqrt{3}-1 AO_1$

Then $P = 3$, $Q = 1$.

The area of $\triangle ABC = \frac{1}{2}AB \cdot AC \cdot \sin A = \frac{1}{2} AB \cdot AC \cdot \sin 60 = \frac{\sqrt{3}}{4} AB \cdot AC$

The area of $\triangle AO_1O_2 = \frac{1}{2} AO_1 \cdot AO_2 \cdot \sin(\angle O_1 A O_2) = \frac{1}{2} AO_1 \cdot (\sqrt{3} -1) AO_1 \cdot \sin(60) = \frac{\sqrt{3}}{4}AO_1^2 (\sqrt{3}-1)$.

$\frac{\text{Area } \triangle ABC}{\text{Area } \triangle AO_1 O_2} = \frac{AB \cdot AC}{(\sqrt{3}-1) AO_1^2} = \frac{\sqrt{3}+1}{2} AC^2 / ( (\sqrt{3}-1)AO_1^2 ) = \frac{(\sqrt{3}+1)}{2}2 AO_1^2 / ( (\sqrt{3}-1)AO_1^2 ) = \frac{(\sqrt{3}+1)}{1} /(\sqrt{3}-1) = (1+\sqrt{3})^2/(3-1) = (1+2\sqrt{3}+3)/2 = (4+2\sqrt{3})/2 = 2+\sqrt{3}$

Thus, $\triangle ABC : \triangle AO_1 O_2 = (2+\sqrt{3})$. $R=4$, $S=2$, $T=3$

3. Final Answer

A = 2

B = 3

C = 2

D = 7

E = 5

F = 1

G = 3

H = 1

I = 3

J = 1

K = 9

L = 0

M = 3

N = 0

O = 2

P = 3

Q = 1

R = 4

S = 2

T = 3