Let $A_1, A_2, \ldots, A_n$ be the vertices of a regular polygon and $A_1A_2$, $A_2A_3$, $\ldots$, $A_{n-1}A_n$, $A_nA_1$ be its $n$ sides. If $\left(\frac{1}{A_1A_2}\right) - \left(\frac{1}{A_1A_4}\right) = \frac{1}{A_1A_3}$, then the value of $n$ is
(a) 5
(b) 6
(c) 7
(d) 8

Suppose $ABCD$ is a quadrilateral such that $\angle BAC = 50^\circ, \angle CAD = 60^\circ, \angle CBD = 30^\circ$ and $\angle BDC = 25^\circ$. If $E$ is the point of intersection of $AC$ and $BD$, then the value of $\angle AEB$ is
(A) $75^\circ$
(B) $85^\circ$
(C) $95^\circ$
(D) $110^\circ$

Suppose $A B C D$ is a quadrilateral such that $\angle B A C = 50 ^ { \circ } , \angle C A D = 60 ^ { \circ } , \angle C B D = 30 ^ { \circ }$ and $\angle B D C = 25 ^ { \circ }$. If $E$ is the point of intersection of $A C$ and $B D$, then the value of $\angle A E B$ is
(A) $75 ^ { \circ }$
(B) $85 ^ { \circ }$
(C) $95 ^ { \circ }$
(D) $110 ^ { \circ }$

Consider a right angled triangle $\triangle A B C$ whose hypotenuse $A C$ is of length 1. The bisector of $\angle A C B$ intersects $A B$ at $D$. If $B C$ is of length $x$, then what is the length of $CD$?
(A) $\sqrt { \frac { 2 x ^ { 2 } } { 1 + x } }$
(B) $\frac { 1 } { \sqrt { 2 + 2 x } }$
(C) $\sqrt { \frac { x } { 1 + x } }$
(D) $\frac { x } { \sqrt { 1 - x ^ { 2 } } }$

In $\triangle ABC$, $CD$ is the median and $BE$ is the altitude. Given that $\overline{CD} = \overline{BE}$, what is the value of $\angle ACD$?
(A) $\pi/3$
(B) $\pi/4$
(C) $\pi/5$
(D) $\pi/6$

Suppose $A B C D$ is a quadrilateral such that $\angle B A C = 50 ^ { \circ } , \angle C A D = 60 ^ { \circ } , \angle C B D = 30 ^ { \circ }$ and $\angle B D C = 25 ^ { \circ }$. If $E$ is the point of intersection of $A C$ and $B D$, then the value of $\angle A E B$ is
(a) $75 ^ { \circ }$.
(B) $85 ^ { \circ }$.
(C) $95 ^ { \circ }$.
(D) $110 ^ { \circ }$.

17. Internal bisector of $\angle \mathrm { A }$ of triangle ABC meets side BC at D . A line drawn through D perpendicular to AD intersects the side AC at E and the side AB at F . If $\mathrm { a } , \mathrm { b } , \mathrm { c }$ represent sides of $\triangle \mathrm { ABC }$ then
(A) AE is HM of b and c
(B) $\mathrm { AD } = \frac { 2 \mathrm { bc } } { \mathrm { b } + \mathrm { c } } \cos \frac { \mathrm { A } } { 2 }$
(C) $\mathrm { EF } = \frac { 4 \mathrm { bc } } { \mathrm { b } + \mathrm { c } } \sin \frac { \mathrm { A } } { 2 }$
(D) the triangle AEF is isosceles
Sol. (A), (B), (C), (D). We have $\triangle \mathrm { ABC } = \triangle \mathrm { ABD } + \triangle \mathrm { ACD }$ $\Rightarrow \quad \frac { 1 } { 2 } \mathrm { bc } \sin \mathrm { A } = \frac { 1 } { 2 } \mathrm { cAD } \sin \frac { \mathrm { A } } { 2 } + \frac { 1 } { 2 } \mathrm {~b} \times \mathrm { AD } \sin \frac { \mathrm { A } } { 2 }$ $\Rightarrow \quad \mathrm { AD } = \frac { 2 \mathrm { bc } } { \mathrm { b } + \mathrm { c } } \cos \frac { \mathrm { A } } { 2 }$ Again $\mathrm { AE } = \mathrm { AD } \sec \frac { \mathrm { A } } { 2 }$ $= \frac { 2 b c } { b + c } \Rightarrow A E$ is HM of $b$ and $c$. [Figure] $\mathrm { EF } = \mathrm { ED } + \mathrm { DF } = 2 \mathrm { DE } = 2 \times \mathrm { AD } \tan \frac { \mathrm { A } } { 2 } = \frac { 2 \times 2 \mathrm { bc } } { \mathrm { b } + \mathrm { c } } \times \cos \frac { \mathrm { A } } { 2 } \times \tan \frac { \mathrm { A } } { 2 }$ $= \frac { 4 \mathrm { bc } } { \mathrm { b } + \mathrm { c } } \sin \frac { \mathrm { A } } { 2 }$ As $\mathrm { AD } \perp \mathrm { EF }$ and $\mathrm { DE } = \mathrm { DF }$ and AD is bisector ⇒ AEF is isosceles. Hence A, B, C and D are correct answers.

In a triangle $XYZ$, let $x, y, z$ be the lengths of sides opposite to the angles $X, Y, Z$, respectively, and $2s = x + y + z$. If $\frac{s-x}{4} = \frac{s-y}{3} = \frac{s-z}{2}$ and area of incircle of the triangle $XYZ$ is $\frac{8\pi}{3}$, then
(A) area of the triangle $XYZ$ is $6\sqrt{6}$
(B) the radius of circumcircle of the triangle $XYZ$ is $\frac{35}{6}\sqrt{6}$
(C) $\sin\frac{X}{2}\sin\frac{Y}{2}\sin\frac{Z}{2} = \frac{4}{35}$
(D) $\sin^2\left(\frac{X+Y}{2}\right) = \frac{3}{5}$

In a triangle $P Q R$, let $\angle P Q R = 30 ^ { \circ }$ and the sides $P Q$ and $Q R$ have lengths $10 \sqrt { 3 }$ and 10, respectively. Then, which of the following statement(s) is (are) TRUE?
(A) $\angle Q P R = 45 ^ { \circ }$
(B) The area of the triangle $P Q R$ is $25 \sqrt { 3 }$ and $\angle Q R P = 120 ^ { \circ }$
(C) The radius of the incircle of the triangle $P Q R$ is $10 \sqrt { 3 } - 15$
(D) The area of the circumcircle of the triangle $P Q R$ is $100 \pi$

In a non-right-angled triangle $\triangle P Q R$, let $p , q , r$ denote the lengths of the sides opposite to the angles at $P , Q , R$ respectively. The median from $R$ meets the side $P Q$ at $S$, the perpendicular from $P$ meets the side $Q R$ at $E$, and $R S$ and $P E$ intersect at $O$. If $p = \sqrt { 3 } , q = 1$, and the radius of the circumcircle of the $\triangle P Q R$ equals 1, then which of the following options is/are correct?
(A) Length of $R S = \frac { \sqrt { 7 } } { 2 }$
(B) Area of $\triangle S O E = \frac { \sqrt { 3 } } { 12 }$
(C) Length of $O E = \frac { 1 } { 6 }$
(D) Radius of incircle of $\triangle P Q R = \frac { \sqrt { 3 } } { 2 } ( 2 - \sqrt { 3 } )$

In a triangle $ABC$, let $AB = \sqrt{23}$, $BC = 3$ and $CA = 4$. Then the value of $$\frac{\cot A + \cot C}{\cot B}$$ is ____.

Let $P Q R S$ be a quadrilateral in a plane, where $Q R = 1 , \angle P Q R = \angle Q R S = 70 ^ { \circ } , \angle P Q S = 15 ^ { \circ }$ and $\angle P R S = 40 ^ { \circ }$. If $\angle R P S = \theta ^ { \circ } , P Q = \alpha$ and $P S = \beta$, then the interval(s) that contain(s) the value of $4 \alpha \beta \sin \theta ^ { \circ }$ is/are
(A) $( 0 , \sqrt { 2 } )$
(B) $( 1,2 )$
(C) $( \sqrt { 2 } , 3 )$
(D) $( 2 \sqrt { 2 } , 3 \sqrt { 2 } )$

Consider an obtuse angled triangle $ABC$ in which the difference between the largest and the smallest angle is $\frac { \pi } { 2 }$ and whose sides are in arithmetic progression. Suppose that the vertices of this triangle lie on a circle of radius 1.
Let $a$ be the area of the triangle $ABC$. Then the value of $( 64a ) ^ { 2 }$ is

ABCD is a trapezium such that AB and CD are parallel and $BC \perp CD$. If $\angle ADB = \theta$, $BC = p$ and $CD = q$, then $AB$ is equal to
(1) $\frac{(p^{2}+q^{2})\sin\theta}{p\cos\theta+q\sin\theta}$
(2) $\frac{p^{2}+q^{2}\cos\theta}{p\cos\theta+q\sin\theta}$
(3) $\frac{p^{2}+q^{2}}{p^{2}\cos\theta+q^{2}\sin\theta}$
(4) $\frac{(p^{2}+q^{2})\sin\theta}{(p\cos\theta+q\sin\theta)^{2}}$

$ABCD$ is a trapezium such that $AB$ and $CD$ are parallel and $BC \perp CD$. If $\angle ADB = \theta$, $BC = p$ and $CD = q$, then $AB$ is equal to
(1) $\frac{p^2 + q^2}{p^2 \cos\theta + q^2 \sin\theta}$
(2) $\frac{\left(p^2 + q^2\right)\sin\theta}{(p\cos\theta + q\sin\theta)^2}$
(3) $\frac{\left(p^2 + q^2\right)\sin\theta}{p\cos\theta + q\sin\theta}$
(4) $\frac{p^2 + q^2\cos\theta}{p\cos\theta + q\sin\theta}$

For a triangle $ABC$, the value of $\cos 2A + \cos 2B + \cos 2C$ is least. If its inradius is 3 and incentre is $M$, then which of the following is NOT correct?
(1) Perimeter of $\triangle ABC$ is $18\sqrt{3}$
(2) $\sin 2A + \sin 2B + \sin 2C = \sin A + \sin B + \sin C$
(3) $\overrightarrow{MA} \cdot \overrightarrow{MB} = -18$
(4) area of $\triangle ABC$ is $\frac{27\sqrt{3}}{2}$

Consider a triangle ABC where
$$\mathrm { AB } = 10 , \quad \angle \mathrm { B } = 30 ^ { \circ }$$
and the radius of its inscribed circle O is 1.
(1) Set $a = \mathrm { BC }$ and $b = \mathrm { CA }$. Finding the area $S$ of the triangle ABC by two different methods, we have
$$S = \frac { \mathbf { A } } { \mathbf{B} } \mathbf { B } ,$$
and
$$S = \frac { \mathbf { C } } { \mathbf{C} } \mathbf { D } ( a + b + \mathbf { E C } ) .$$
Hence we obtain
$$b = \mathbf { G } a - \mathbf { H I } .$$
Since the relationship between $a$ and $b$ can also be expressed by the equation
$$b ^ { 2 } = a ^ { 2 } - \mathbf { J K } \sqrt { \mathbf { L } } a + \mathbf { M N O } ,$$
we have
$$a = \frac { \mathbf { P Q } - \mathbf { R } } { 3 } \sqrt { \mathbf { S } } , \quad b = \frac { \mathbf { T U } } { \mathbf{P} } \cdot \frac { \mathbf { V } } { 3 } \sqrt { \mathbf { W } } .$$
(2) Let D denote the point of intersection of the segment BC and the straight line which passes through the two points A and O. We denote the area of the triangle OBC by $S ^ { \prime }$. Since
$$S : S ^ { \prime } = \mathbf { X } : 1 ,$$
it follows that
$$\mathrm { AO } : \mathrm { OD } = \mathbf { Y } : 1 .$$

Consider a triangle ABC where $\mathrm{AB} = 8$, $\mathrm{AC} = 5$ and $\angle\mathrm{BAC} = 60^\circ$. Take point D and point E on sides AB and AC, respectively, such that the segment DE divides triangle ABC into two parts with the equal areas. Set $x = \mathrm{AD}$.
(1) When we represent AE in terms of $x$, we have $\mathrm{AE} = \frac{\mathbf{JK}}{x}$.
(2) When E moves along side AC, the range of the values which $x$ can take is
$$\mathbf{L} \leq x \leqq \mathbf{M}.$$
(3) Since $\mathrm{DE}^2 = \left(x - \frac{\mathbf{NO}}{x}\right)^2 + \mathbf{PQ}$, the length of segment DE is minimized at $x = \mathbf{R}\sqrt{\mathbf{S}}$, and its length there is $\mathbf{T}$.$\mathbf{U}$.

Suppose that in the figure to the right
$$\mathrm { AB } = 4 , \quad \mathrm { AC } = 5 , \quad \cos \angle \mathrm { BAC } = \frac { 1 } { 8 }$$
and
$$\angle \mathrm { BAD } = \angle \mathrm { ACB } , \quad \angle \mathrm { CAE } = \angle \mathrm { ABC } .$$
(1) When we denote the area of $\triangle \mathrm { ABC }$ by $S$, we have
$$S = \frac { \square \mathbf { A B } \sqrt { \square \mathbf { C } } } { \square } .$$
Also $\mathrm { BC } = \mathbf { E }$.
(2) Furthermore, when we denote the areas of $\triangle \mathrm { ABD }$ and $\triangle \mathrm { ACE }$ by $S _ { 1 }$ and $S _ { 2 }$, respectively, we have
$$S : S _ { 1 } : S _ { 2 } = 1 : \frac { \mathbf { F } } { \mathbf{G} } : \frac { \mathbf { H I } } { \mathbf { J } } .$$
(3) When we denote the area of $\triangle \mathrm { ADE }$ by $T$, we have
$$T = \frac { \mathbf { L M } \sqrt { \mathbf { N } } } { \mathbf { O P } } .$$
Also $\mathrm { DE } = \dfrac { \mathbf { Q } } { \mathbf{R} }$.

In the figure to the right, let $$\mathrm{OA} = 6, \quad \mathrm{OB} = 3, \quad \angle\mathrm{AOB} = 120^\circ,$$ and let the point Q denote the point of intersection of the bisector of $\angle\mathrm{XAB}$ and the bisector of $\angle\mathrm{ABY}$. Let P denote the point of intersection of segment AB and segment OQ. We are to find the length of segment PQ.
(1) First of all, we see that $\mathrm{AB} = \square\sqrt{\mathbf{A}}$ and that the area of triangle OAB is $\mathbf{BC}\sqrt{\mathbf{D}}$.
(2) For $\mathbf{F}$ and $\mathbf{G}$ in the following, choose the correct answer from among choices (0) $\sim$ (4), just below. (0) AB (1) AP (2) AQ (3) BP (4) BQ
Since AQ is the bisector of the exterior angle of $\angle\mathrm{A}$ of triangle OAP and BQ is the bisector of the exterior angle of $\angle\mathrm{B}$ of triangle OBP, we have $$\begin{aligned} \mathrm{OQ} : \mathrm{PQ} &= \mathrm{OA} : \mathbf{F} \\ &= \mathrm{OB} : \mathbf{G} \end{aligned}$$
(3) Thus we see $\mathrm{AP} = \mathbf{H}\sqrt{\mathbf{I}}$. Since $\angle\mathrm{AOP} = \mathbf{JK}^\circ$, we have $\mathrm{OP} = \square\mathbf{L}$. Hence we have $\mathrm{PQ} = \mathbf{M} + \mathbf{N}\sqrt{\mathbf{O}}$.

Consider a triangle ABC where $\angle \mathrm{ BAC } = 60 ^ { \circ }$.
Let D be the point of intersection of the bisector of $\angle \mathrm{ BAC }$ and the side BC. Let DE and DF be the line segments perpendicular to sides AB and AC, respectively. Let us set
$$x = \frac { \mathrm{ AB } } { \mathrm{ AC } } , \quad k = \frac { \triangle \mathrm{ DEF } } { \triangle \mathrm{ ABC } } .$$
Note that $\triangle \mathrm{ ABC }$ denotes the area of the triangle ABC, and similarly for other triangles.
(1) We are to represent $k$ in terms of $x$. Since $\triangle \mathrm{ ABD } + \triangle \mathrm{ ACD } = \triangle \mathrm{ ABC }$, when we set $b = \mathrm{ AB }$, $c = \mathrm{ AC }$ and $d = \mathrm{ AD }$, we have
$$d = \frac { \sqrt { \mathbf { A } } \, b c } { b + c } .$$
Next, since $\mathrm{ DE } = \mathrm{ DF } = \dfrac { \mathbf { B } } { \mathbf { C } } d$, we have
$$\triangle \mathrm{ DEF } = \frac { \sqrt { \mathbf { D } } } { \mathbf { EF } } d ^ { 2 } .$$
From (1) and (2), we see that
$$k = \frac { d ^ { 2 } } { \mathbf { G } \, b c } = \frac { \mathbf { H } \, b c } { \mathbf { I } ( b + c ) ^ { 2 } } .$$
Since $x = \dfrac { b } { c }$, we have
$$k = \frac { \mathbf { J } \, x } { \mathbf { K } ( x + \mathbf { L } ) ^ { 2 } } .$$
(2) If $\mathrm{ BD } = 8$ and $\mathrm{ BC } = 10$, then $x = \mathbf { M }$ and $k = \dfrac { \mathbf { N } } { \mathbf { O P } }$.

[1] In $\triangle \mathrm { ABC }$, $\mathrm { AB } = \sqrt { 3 } - 1$, $\mathrm { BC } = \sqrt { 3 } + 1$, and $\angle \mathrm { ABC } = 60 ^ { \circ }$.
(1) $\mathrm { AC } = \sqrt { \text { A } }$, so the circumradius of $\triangle \mathrm { ABC }$ is $\sqrt { \text { B } }$, and
$$\sin \angle \mathrm { BAC } = \frac { \sqrt { \square } + \sqrt { \square } } { \square }$$
Here, the order of answers for □ C and □ D does not matter.
(2) Let D be a point on side AC such that the area of $\triangle \mathrm { ABD }$ is $\frac { \sqrt { 2 } } { 6 }$. Then
$$\mathrm { AB } \cdot \mathrm { AD } = \frac { \square \sqrt { * } - \square } { \square }$$
Therefore, $\mathrm { AD } = \frac { \square } { \square }$. □ E F
[2] Ski jumping is a competition where athletes compete on both jump distance and the beauty of their aerial posture. Athletes slide down a slope and take off from the edge of the slope into the air. A score $X$ is determined from the jump distance $D$ (in meters), and a score $Y$ is determined from the aerial posture. Consider 58 jumps at a certain competition.
(1) For the score $X$, the score $Y$, and the takeoff velocity $V$ (in km/h), three scatter plots in Figure 1 were obtained.
Choose one from options (0)–(6) below for each of G, H, and I. The order of answers does not matter.
The correct statements that can be read from Figure 1 are □ G, □ H, and □ I. (0) The correlation between $X$ and $V$ is stronger than the correlation between $X$ and $Y$.
(1) There is a positive correlation between $X$ and $Y$.
(2) The jump with maximum $V$ also has maximum $X$.
(3) The jump with maximum $V$ also has maximum $Y$.
(4) The jump with minimum $Y$ does not have minimum $X$.
(5) All jumps with $X \geq 80$ have $V \geq 93$. (6) There is no jump with $Y \geq 55$ and $V \geq 94$.
(2) The score $X$ is calculated from the jump distance $D$ using the following formula:
$$X = 1.80 \times ( D - 125.0 ) + 60.0$$
Choose one from options (0)–(6) below for each of J, K, and L. You may select the same option more than once. – The variance of $X$ is □ J times the variance of $D$. – The covariance of $X$ and $Y$ is □ K times the covariance of $D$ and $Y$. Here, covariance is defined as the average of the products of deviations from the mean for each of the two variables. – The correlation coefficient of $X$ and $Y$ is □ L times the correlation coefficient of $D$ and $Y$. (0) $- 125$
(1) $- 1.80$
(2) $1$
(3) $1.80$
(4) $3.24$
(5) $3.60$ (6) $60.0$
(3) The 58 jumps were performed by 29 athletes, each performing 2 jumps. A histogram for the values of $X + Y$ (sum of scores $X$ and $Y$) for the first jump and a histogram for the values of $X + Y$ for the second jump are either A or B in Figure 2. Also, a box plot for the values of $X + Y$ for the first jump and a box plot for the values of $X + Y$ for the second jump are either a or b in Figure 3. The minimum value of $X + Y$ for the first jump was 108.0.
Choose one from options ⓪–③ in the table below for the blank M.
The correct combination of histogram and box plot for the values of $X + Y$ for the first jump is □ M.

\cline { 2 - 5 } \multicolumn{1}{c|}{}	(0)	(1)	(2)	(3)
Histogram	A	A	B	B
Box plot	a	b	a	b

Choose one from options (0)–(3) below for the blank N.
The correct statement that can be read from Figure 3 is □ N. (0) The interquartile range of $X + Y$ for the first jump is larger than the interquartile range of $X + Y$ for the second jump.
(1) The median of $X + Y$ for the first jump is larger than the median of $X + Y$ for the second jump.
(2) The maximum value of $X + Y$ for the first jump is smaller than the maximum value of $X + Y$ for the second jump.
(3) The minimum value of $X + Y$ for the first jump is smaller than the minimum value of $X + Y$ for the second jump.

The triangle ABC satisfies
$$\mathrm { AB } = 4 , \quad \mathrm { AC } = 3 \quad \text { and } \quad \angle \mathrm { B } = 30 ^ { \circ } .$$
D is the point on side BC such that $\mathrm { AC } = \mathrm { AD }$. Let us consider the circumscribed circle O of triangle ACD.
(1) Since $\sin B = \frac { \mathbf { A } } { \mathbf { B } }$, we have $\sin C = \frac { \mathbf { C } } { \mathbf{D} }$.
Hence the radius of circle O is $\frac { \mathbf { E } } { \mathbf{F} }$.
(2) We have
$$\mathrm { BC } = \mathrm { G } \sqrt { \mathrm { H } } + \sqrt { \mathrm{H} }$$
and
$$\mathrm { BD } = \mathrm { J } \sqrt { \mathrm {~K} } - \sqrt { \mathrm { K } } .$$
Let us denote the intersection of side AB and circle O by E . Then
$$\mathrm { BE } = \frac { \mathbf { M } } { \mathbf{N} } .$$
Hence the relationships between the areas of triangles $\mathrm { BDE } , \mathrm { ADE }$ and ACD are
$$\begin{aligned} & \triangle \mathrm { BDE } : \triangle \mathrm { ADE } = \mathbf { O } : \mathbf { P } , \\ & \triangle \mathrm { BDE } : \triangle \mathrm { ACD } = \mathbf { Q } ( \mathbf{J} \sqrt { \mathbf { K } } - \sqrt { \mathrm { K } } ) : \mathbf { R S } \sqrt { \mathbf { T } } . \end{aligned}$$

Consider a triangle ABC and its circumscribed circle O, where the lengths of the three sides of the triangle are $$\mathrm{AB} = 2, \quad \mathrm{BC} = 3, \quad \mathrm{CA} = 4.$$ Below, the area of a triangle such as PQR is expressed as $\triangle\mathrm{PQR}$.
(1) We see that $\cos\angle\mathrm{ABC} = \frac{\mathbf{AB}}{\mathbf{C}}$.
(2) Let us take a point D on the circumference of circle O such that it is on the opposite side of the circle from point B with respect to AC and $$\frac{\triangle\mathrm{ABD}}{\triangle\mathrm{BCD}} = \frac{8}{15}.$$ We are to find the lengths of line segments AD and CD.
First, since $$\angle\mathrm{BAD} = \mathbf{DEF}^\circ - \angle\mathrm{BCD},$$ we have $\sin\angle\mathrm{BAD} = \sin\angle\mathrm{BCD}$. Hence from (1) we have $$\frac{\mathrm{AD}}{\mathrm{CD}} = \frac{\mathbf{G}}{\mathbf{H}},$$ so we set $\mathrm{AD} = \mathbf{G}k$ and $\mathrm{CD} = \mathbf{H}k$, where $k$ is a positive number. Furthermore, since $$\angle\mathrm{ADC} = \mathbf{IJK}^\circ - \angle\mathrm{ABC},$$ we have $\cos\angle\mathrm{ADC} = \frac{\mathbf{L}}{\mathbf{L}}$. Hence, we obtain $k = \frac{\mathbf{N}}{\sqrt{\mathbf{OP}}}$, and then $$\mathrm{AD} = \frac{\mathbf{QR}\sqrt{\mathbf{OP}}}{\mathbf{OP}},$$ $$\mathrm{CD} = \frac{\mathbf{ST}\sqrt{\mathbf{OP}}}{\mathbf{OP}}.$$
(3) When we denote the point of intersection of the straight line DA and the straight line CB by E, we have $$\frac{\triangle\mathrm{ABE}}{\triangle\mathrm{CDE}} = \frac{\mathbf{UV}}{\mathbf{WXY}}.$$

In a triangle ABC, let $\angle \mathrm { B } = 45 ^ { \circ }$ and $\angle \mathrm { C } = 75 ^ { \circ }$, and let D be the intersection of the bisector of $\angle \mathrm { A }$ and side BC.
(1) From the law of sines we have
$$\mathrm { AC } = \frac { \sqrt { \mathbf { A } } } { \sqrt { \mathbf { B } } } \mathrm { BC } , \quad \mathrm { AD } = \sqrt { \mathbf { C } } \mathrm { BD } .$$
In particular, from $\angle \mathrm { ADC } = \mathbf { D E } ^ { \circ }$ we see that
$$\mathrm { BD } : \mathrm { BC } = \mathbf { F } : \sqrt { \mathbf { G } }$$
and hence we have
$$\mathrm { AB } : \mathrm { AC } = \mathbf { H } : \left( \sqrt { \mathbf { I } } - \frac { \mathbf { J } }{\mathbf{J} } \right) .$$
(2) Let $\mathrm { O } _ { 1 }$ be the center of the circumscribed circle of triangle ABD, and let $\mathrm { O } _ { 2 }$ be the center of the circumscribed circle of triangle ADC. Let us find the ratio of the areas of triangle ABC and triangle $\mathrm { AO } _ { 1 } \mathrm { O } _ { 2 }$, $\triangle \mathrm { ABC } : \triangle \mathrm { AO } _ { 1 } \mathrm { O } _ { 2 }$.
Since $\angle \mathrm { AO } _ { 1 } \mathrm { D } = \mathbf { K } \mathbf { L } ^ { \circ }$ and $\angle \mathrm { AO } _ { 2 } \mathrm { O } _ { 1 } = \mathbf { M N } ^ { \circ }$, by the same reasoning as (1), we have
$$\mathrm { AC } = \sqrt { \mathbf { O } } \mathrm { AO } _ { 1 } , \quad \mathrm { AO } _ { 2 } = ( \sqrt { \mathbf { P } } - \mathbf { Q } ) \mathrm { AO } _ { 1 } .$$
Hence we obtain
$$\triangle \mathrm { ABC } : \triangle \mathrm { AO } _ { 1 } \mathrm { O } _ { 2 } = \mathbf { R } : ( \mathbf { S } - \sqrt { \mathbf { T } } ) .$$