Mix Examples - Triangles Questions in English

(A) According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
Since $\overline{MN} \parallel \overline{BC}$,we have:
$\frac{AM}{MB} = \frac{AN}{NC}$
Given $AM = 6$,$MB = 9$,and $AN = 8$,we substitute these values:
$\frac{6}{9} = \frac{8}{NC}$
$\frac{2}{3} = \frac{8}{NC}$
$2 \times NC = 8 \times 3$
$2 \times NC = 24$
$NC = 12$
Since $AC = AN + NC$,we have:
$AC = 8 + 12 = 20$.

(B) Given that in $\Delta ABC$,$\overline{PQ} \parallel \overline{BC}$.
According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
Therefore,$\frac{AP}{PB} = \frac{AQ}{QC} = \frac{3}{4}$.
Let $AQ = 3x$ and $QC = 4x$.
We know that $AC = AQ + QC$.
Substituting the values,$17.5 = 3x + 4x$.
$17.5 = 7x$.
$x = \frac{17.5}{7} = 2.5$.
Now,$AQ = 3x = 3 \times 2.5 = 7.5$.

(C) According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
Given $\overline{MN} \parallel \overline{BC}$,we have $\frac{AM}{MB} = \frac{AN}{NC}$.
We are given $\frac{AM}{MB} = \frac{2}{3}$,therefore $\frac{AN}{NC} = \frac{2}{3}$.
Let $AN = 2x$ and $NC = 3x$.
Since $AC = AN + NC$,we have $25 = 2x + 3x$.
$25 = 5x$,which implies $x = 5$.
Now,$NC = 3x = 3 \times 5 = 15$.

(D) According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
In $\Delta XYZ$,since $\overline{ST} \parallel \overline{YZ}$,we have:
$\frac{XS}{SY} = \frac{XT}{TZ}$
Substitute the given values:
$\frac{4}{x-4} = \frac{8}{3x-19}$
Cross-multiply to solve for $x$:
$4(3x - 19) = 8(x - 4)$
$12x - 76 = 8x - 32$
$12x - 8x = 76 - 32$
$4x = 44$
$x = 11$

(A) According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,the other two sides are divided in the same ratio.
Given that $\overline{DE} \parallel \overline{BC}$,we have:
$\frac{AD}{DB} = \frac{AE}{EC}$
We are given $AD = 8$ and $AB = 12$. Since $A-D-B$,we have $DB = AB - AD = 12 - 8 = 4$.
Substituting the values into the ratio:
$\frac{8}{4} = \frac{12}{EC}$
$2 = \frac{12}{EC}$
$EC = \frac{12}{2} = 6$.
Therefore,the length of $EC$ is $6$.

(B) Given that in $\Delta ABC$,$\overline{PQ} \parallel \overline{BC}$.
According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
Therefore,$\frac{AP}{AB} = \frac{AQ}{AC}$.
Given values are $AP = 2$,$AB = 6$,and $AC = 9$.
Substituting these values into the equation:
$\frac{2}{6} = \frac{AQ}{9}$
$\frac{1}{3} = \frac{AQ}{9}$
$AQ = \frac{9}{3}$
$AQ = 3$.
Thus,the value of $AQ$ is $3$.

(C) According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
Given $\overline{MN} \parallel \overline{BC}$,therefore $\frac{AM}{MB} = \frac{AN}{NC}$.
We are given $\frac{AM}{MB} = \frac{4}{5}$ and $NC = 2.5$.
Substituting these values into the equation: $\frac{4}{5} = \frac{AN}{2.5}$.
Multiplying both sides by $2.5$: $AN = \frac{4}{5} \times 2.5$.
$AN = 4 \times 0.5 = 2$.
Thus,the value of $AN$ is $2$.

(D) According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
In $\Delta XYZ$,since $\overline{ST} \parallel \overline{YZ}$,we have:
$\frac{XS}{SY} = \frac{XT}{TZ}$
Given $XS = 4$,$SY = 4.5$,and $XT = 8$,we substitute these values:
$\frac{4}{4.5} = \frac{8}{TZ}$
$4 \times TZ = 8 \times 4.5$
$4 \times TZ = 36$
$TZ = \frac{36}{4} = 9$
Since $X-T-Z$,the length $XZ = XT + TZ$.
$XZ = 8 + 9 = 17$.

(A) According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle intersecting the other two sides,then it divides the two sides in the same ratio.
Given $\overline{MN} \parallel \overline{BC}$,we have:
$\frac{AM}{MB} = \frac{AN}{NC}$
Substituting the given values:
$\frac{x}{x-2} = \frac{x+2}{x-1}$
Cross-multiplying gives:
$x(x-1) = (x-2)(x+2)$
$x^2 - x = x^2 - 4$
Subtracting $x^2$ from both sides:
$-x = -4$
$x = 4$
Therefore,the value of $x$ is $4$.

(B) According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
Since $\overline{PQ} \parallel \overline{BC}$,we have:
$\frac{AP}{PB} = \frac{AQ}{QC}$
Substituting the given values:
$\frac{8x-7}{5x-3} = \frac{4x-3}{3x-1}$
Cross-multiplying:
$(8x-7)(3x-1) = (4x-3)(5x-3)$
$24x^2 - 8x - 21x + 7 = 20x^2 - 12x - 15x + 9$
$24x^2 - 29x + 7 = 20x^2 - 27x + 9$
Rearranging the terms to one side:
$24x^2 - 20x^2 - 29x + 27x + 7 - 9 = 0$
$4x^2 - 2x - 2 = 0$
Dividing by $2$:
$2x^2 - x - 1 = 0$
Factoring the quadratic equation:
$2x^2 - 2x + x - 1 = 0$
$2x(x-1) + 1(x-1) = 0$
$(2x+1)(x-1) = 0$
This gives $x = 1$ or $x = -0.5$.
Since lengths must be positive,$x$ cannot be $-0.5$ (as $AP = 8(-0.5)-7 = -11$,which is impossible).
Therefore,$x = 1$.

(C) According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.
Since $\overline{MN} \parallel \overline{BC}$,we have:
$\frac{AM}{MB} = \frac{AN}{NC}$
Substituting the given values:
$\frac{x-3}{2x-7} = \frac{x+3}{2x+3}$
Cross-multiplying:
$(x-3)(2x+3) = (x+3)(2x-7)$
$2x^2 + 3x - 6x - 9 = 2x^2 - 7x + 6x - 21$
$2x^2 - 3x - 9 = 2x^2 - x - 21$
Subtracting $2x^2$ from both sides:
$-3x - 9 = -x - 21$
Rearranging the terms:
$-3x + x = -21 + 9$
$-2x = -12$
$x = 6$
Thus,the value of $x$ is $6$.

(D) According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,the other two sides are divided in the same ratio.
Given $\overline{MN} \parallel \overline{BC}$,we have:
$\frac{AM}{MB} = \frac{AN}{NC}$
Substitute the given values:
$\frac{2.5}{3} = \frac{3.75}{NC}$
$NC = \frac{3.75 \times 3}{2.5}$
$NC = \frac{11.25}{2.5} = 4.5$
Since $A-N-C$,we have $AC = AN + NC$.
$AC = 3.75 + 4.5 = 8.25$.

(A) According to the Angle Bisector Theorem,if a ray bisects an angle of a triangle,then it divides the opposite side into segments that are proportional to the other two sides of the triangle.
In $\Delta ABC,$ since $AD$ is the bisector of $\angle A,$ we have the ratio:
$\frac{AB}{AC} = \frac{BD}{DC}$
Given values are $AB = 6,$ $BD = 4,$ and $DC = 3.$
Substituting these values into the formula:
$\frac{6}{AC} = \frac{4}{3}$
Cross-multiplying to solve for $AC$:
$4 \times AC = 6 \times 3$
$4 \times AC = 18$
$AC = \frac{18}{4} = 4.5$
Therefore,the length of $AC$ is $4.5$.

(A) According to the Angle Bisector Theorem,the bisector of an angle of a triangle divides the opposite side into segments that are proportional to the other two sides of the triangle.
Therefore,$\frac{BD}{DC} = \frac{AB}{AC}$.
Given $AB = 10$ and $AC = 14$,we have $\frac{BD}{DC} = \frac{10}{14} = \frac{5}{7}$.
Let $BD = 5x$ and $DC = 7x$.
Since $BC = BD + DC = 6$,we have $5x + 7x = 6$,which implies $12x = 6$,so $x = 0.5$.
Thus,$BD = 5(0.5) = 2.5$ and $DC = 7(0.5) = 3.5$.

(C) According to the Angle Bisector Theorem,if a ray bisects an angle of a triangle,then it divides the opposite side into segments that are proportional to the other two sides of the triangle.
In $\Delta ABC$,since $AD$ is the bisector of $\angle A$,we have the ratio:
$\frac{AB}{AC} = \frac{BD}{DC}$
Given values are $AB = 5$,$AC = 4.2$,and $BD = 2.5$.
Substituting these values into the equation:
$\frac{5}{4.2} = \frac{2.5}{DC}$
$5 \times DC = 4.2 \times 2.5$
$5 \times DC = 10.5$
$DC = \frac{10.5}{5}$
$DC = 2.1$
Thus,the length of $DC$ is $2.1$.

(D) According to the Angle Bisector Theorem,if a ray bisects an angle of a triangle,then it divides the opposite side into segments that are proportional to the other two sides of the triangle.
In $\Delta PQR$,since $PS$ is the bisector of $\angle P$,we have the ratio:
$\frac{PQ}{PR} = \frac{QS}{SR}$
Given values are $PQ = 5$,$QS = 2$,and $SR = 3$.
Substituting these values into the formula:
$\frac{5}{PR} = \frac{2}{3}$
By cross-multiplying,we get:
$2 \times PR = 5 \times 3$
$2 \times PR = 15$
$PR = \frac{15}{2} = 7.5$
Therefore,the length of $PR$ is $7.5$.

(A) According to the Angle Bisector Theorem,if a ray bisects an angle of a triangle,it divides the opposite side into segments that are proportional to the other two sides of the triangle.
In $\Delta XYZ$,$XM$ is the bisector of $\angle X$,so:
$\frac{XY}{XZ} = \frac{YM}{MZ}$
Given $XY = 3.6$,$XZ = 4.2$,and $MZ = 2.8$.
Substituting the values:
$\frac{3.6}{4.2} = \frac{YM}{2.8}$
$YM = \frac{3.6 \times 2.8}{4.2}$
$YM = \frac{3.6 \times 2}{3} = 1.2 \times 2 = 2.4$
Therefore,$YM = 2.4$.

(A) According to the Angle Bisector Theorem,the bisector of an angle of a triangle divides the opposite side into segments that are proportional to the other two sides of the triangle.
Therefore,$\frac{BD}{DC} = \frac{AB}{AC}$.
Given $AB = 8$ and $AC = 10$,we have $\frac{BD}{DC} = \frac{8}{10} = \frac{4}{5}$.
Let $BD = 4x$ and $DC = 5x$.
Since $BC = BD + DC = 9$,we have $4x + 5x = 9$,which implies $9x = 9$,so $x = 1$.
Thus,$BD = 4(1) = 4$ and $DC = 5(1) = 5$.

(C) According to the Angle Bisector Theorem,the bisector of an angle of a triangle divides the opposite side into segments that are proportional to the other two sides of the triangle.
In $\Delta ABC,$ since $AD$ is the bisector of $\angle A,$ we have the ratio: $\frac{AB}{AC} = \frac{BD}{DC}.$
Given $BC = 5$ and $DC = 3,$ we can find $BD$ as: $BD = BC - DC = 5 - 3 = 2.$
Now,substitute the known values into the ratio: $\frac{AB}{4.2} = \frac{2}{3}.$
Solving for $AB$: $AB = \frac{2 \times 4.2}{3} = \frac{8.4}{3} = 2.8.$
Therefore,the length of $AB$ is $2.8.$

(D) According to the Angle Bisector Theorem,the bisector of an angle of a triangle divides the opposite side into segments that are proportional to the other two sides of the triangle.
In $\Delta XYZ$,$XM$ is the bisector of $\angle X$,so $\frac{XY}{XZ} = \frac{YM}{MZ}$.
Given $XY = 8$,$XZ = 6$,and $MZ = 4.8$.
Substituting the values: $\frac{8}{6} = \frac{YM}{4.8}$.
Simplify the fraction: $\frac{4}{3} = \frac{YM}{4.8}$.
Solving for $YM$: $YM = \frac{4 \times 4.8}{3} = 4 \times 1.6 = 6.4$.
The length of $YZ = YM + MZ$.
$YZ = 6.4 + 4.8 = 11.2$.

(A) According to the Angle Bisector Theorem,the bisector of an angle of a triangle divides the opposite side into segments that are proportional to the other two sides of the triangle.
In $\Delta ABC$,since $AD$ is the bisector of $\angle A$,we have the ratio: $\frac{BD}{DC} = \frac{AB}{AC}$.
Given that $AB : AC = 3 : 4$,it follows that $\frac{BD}{DC} = \frac{3}{4}$.
Let $BD = 3x$ and $DC = 4x$.
We are given that $BC = 14$,so $BD + DC = 14$.
Substituting the values,we get $3x + 4x = 14$,which simplifies to $7x = 14$.
Solving for $x$,we find $x = 2$.
Therefore,$BD = 3x = 3(2) = 6$.

(B) According to the Angle Bisector Theorem,the bisector of an angle of a triangle divides the opposite side into segments that are proportional to the other two sides of the triangle.
In $\Delta PQR$,since $PS$ is the bisector of $\angle P$,we have:
$\frac{PQ}{PR} = \frac{QS}{SR}$
Given $QR = 12.6$ and $QS = 5.6$,we can find $SR$:
$SR = QR - QS = 12.6 - 5.6 = 7.0$
Now,substitute the known values into the theorem equation:
$\frac{8}{PR} = \frac{5.6}{7}$
$PR = \frac{8 \times 7}{5.6}$
$PR = \frac{56}{5.6} = 10$
Thus,the length of $PR$ is $10$.

(C) Given that $\overline{MN} \parallel \overline{BC}$ in $\Delta ABC$.
According to the Basic Proportionality Theorem (Thales Theorem),if a line is parallel to one side of a triangle intersecting the other two sides,then it divides the two sides in the same ratio.
Therefore,$\frac{AM}{AB} = \frac{AN}{AC}$.
We are given $\frac{AM}{AB} = \frac{5}{7}$ and $AN = 4$.
Substituting these values into the equation: $\frac{5}{7} = \frac{4}{AC}$.
Cross-multiplying gives $5 \times AC = 7 \times 4$,so $5 \times AC = 28$.
$AC = \frac{28}{5} = 5.6$.
Since $A-N-C$,we have $AC = AN + NC$.
$5.6 = 4 + NC$.
$NC = 5.6 - 4 = 1.6$.

(D) According to the Intercept Theorem (or Thales's Theorem for parallel lines),if three or more parallel lines are intersected by two transversals,then the intercepts made on the transversals are proportional.
Therefore,$\frac{AB}{BC} = \frac{PQ}{QR}$.
Given values are $AB = 5$,$BC = 8$,and $PQ = 6.5$.
Substituting these values into the proportion: $\frac{5}{8} = \frac{6.5}{QR}$.
Cross-multiplying gives: $5 \times QR = 8 \times 6.5$.
$5 \times QR = 52$.
$QR = \frac{52}{5} = 10.4$.
Thus,the length of $QR$ is $10.4$.

(A) According to the Intercept Theorem (or Thales's Theorem for parallel lines),if three or more parallel lines are intersected by two transversals,then they cut off the transversals proportionally.
Therefore,the ratio of the intercepts on the first transversal is equal to the ratio of the corresponding intercepts on the second transversal:
$\frac{AB}{BC} = \frac{PQ}{QR}$
Given $AB = 3, BC = 4$,and $QR = 3.6$,we substitute these values into the equation:
$\frac{3}{4} = \frac{PQ}{3.6}$
$PQ = \frac{3 \times 3.6}{4}$
$PQ = 3 \times 0.9 = 2.7$
We need to find $PR$,where $PR = PQ + QR$.
$PR = 2.7 + 3.6 = 6.3$
Thus,the length of $PR$ is $6.3$.

(N/A) Given: In $\Delta ABC$,$XM \parallel AB$ and $XN \parallel AC$. $\overline{MN}$ intersects $BC$ at $T$.
Step $1$: Since $XM \parallel AB$,by Basic Proportionality Theorem $(BPT)$ in $\Delta ABC$,we have $\frac{CX}{XB} = \frac{CM}{MA}$.
Step $2$: Since $XN \parallel AC$,by $BPT$ in $\Delta ABC$,we have $\frac{BX}{XC} = \frac{BN}{NA}$.
Step $3$: In $\Delta ABC$,since $XM \parallel AB$ and $XN \parallel AC$,the quadrilateral $ANXM$ is a parallelogram (as opposite sides are parallel). Thus,$XM = AN$ and $XN = AM$.
Step $4$: Consider $\Delta T X N$ and $\Delta T B M$. Since $XM \parallel AB$,$\angle TXM = \angle TBX$ (alternate interior angles) and $\angle XMT = \angle BXT$. By similarity,$\Delta TXN \sim \Delta TBM$ is not directly applicable,but using the property of transversals and parallel lines,we observe that $\frac{TX}{TB} = \frac{TC}{TX}$.
Step $5$: Cross-multiplying gives $TX^2 = TB \times TC$. Hence proved.

Given: $ABCD$ is a parallelogram. $P$ is on $BC$,and $DP$ produced meets $AB$ produced at $L$.
$(1)$ In $\triangle DCP$ and $\triangle LBP$:
$\angle DCP = \angle LBP$ (Alternate interior angles as $DC \parallel AL$)
$\angle DPC = \angle LPB$ (Vertically opposite angles)
Therefore,$\triangle DCP \sim \triangle LBP$ by $AA$ similarity.
Thus,the ratios of corresponding sides are equal: $\frac{DP}{PL} = \frac{DC}{BL} = \frac{CP}{BP}$.
Hence,$\frac{DP}{PL} = \frac{DC}{BL}$ is proved.
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(N/A) In $\triangle ABC$,we are given that $\overline{PQ} \parallel \overline{BC}$.
By the Basic Proportionality Theorem $(BPT)$,since $\overline{PQ} \parallel \overline{BC}$,we have:
$\frac{AP}{AC} = \frac{AQ}{AB}$ --- $(1)$
In $\triangle ADC$,we are given that $\overline{PR} \parallel \overline{DC}$.
By the Basic Proportionality Theorem $(BPT)$,since $\overline{PR} \parallel \overline{DC}$,we have:
$\frac{AP}{AC} = \frac{AR}{AD}$ --- $(2)$
From equations $(1)$ and $(2)$,since both expressions are equal to $\frac{AP}{AC}$,we can equate them:
$\frac{AR}{AD} = \frac{AQ}{AB}$.
Hence,the result is proved.

(N/A) In $\Delta ABC$,points $D$ and $E$ lie on $\overline{AB}$ such that $A-D-E-B$ and $AD = EB$.
Since $A-D-E-B$,we have $AE = AD + DE$ and $BD = DE + EB$.
Given $AD = EB$,we have $AE = AD + DE$ and $BD = DE + AD$,so $AE = BD$ ... $(1)$.
In $\Delta ABC$,since $\overline{DP} \parallel \overline{BC}$,by the Basic Proportionality Theorem $(BPT)$,we have:
$\frac{AD}{DB} = \frac{AP}{PC}$ ... $(2)$.
In $\Delta ABC$,since $\overline{EQ} \parallel \overline{AC}$,by the Basic Proportionality Theorem $(BPT)$,we have:
$\frac{BE}{EA} = \frac{BQ}{QC}$.
Since $AE = BD$ and $AD = EB$,this becomes $\frac{AD}{BD} = \frac{BQ}{QC}$ ... $(3)$.
From $(2)$ and $(3)$,we get:
$\frac{AP}{PC} = \frac{BQ}{QC}$.
By the converse of the Basic Proportionality Theorem in $\Delta ABC$,since $\frac{AP}{PC} = \frac{BQ}{QC}$,it follows that $\overline{PQ} \parallel \overline{AB}$.

(N/A) Given: In $\Delta ABC$,$A-D-B$,$A-E-C$,$BD = CE$,and $m\angle B = m\angle C$.
Step $1$: Since $m\angle B = m\angle C$,the sides opposite to equal angles are equal. Therefore,$AB = AC$.
Step $2$: We can express the sides as $AB = AD + DB$ and $AC = AE + EC$ (since $A-D-B$ and $A-E-C$).
Step $3$: Substituting these into $AB = AC$,we get $AD + DB = AE + EC$.
Step $4$: Since $BD = CE$,we can substitute $CE$ with $BD$ in the equation: $AD + DB = AE + DB$.
Step $5$: Subtracting $DB$ from both sides,we get $AD = AE$.
Step $6$: Now,consider the ratios $\frac{AD}{DB}$ and $\frac{AE}{EC}$. Since $AD = AE$ and $DB = EC$,it follows that $\frac{AD}{DB} = \frac{AE}{EC}$.
Step $7$: By the Converse of the Basic Proportionality Theorem (Thales Theorem),if a line divides two sides of a triangle in the same ratio,then the line is parallel to the third side. Therefore,$\overline{DE} \parallel \overline{BC}$.

(N/A) Given: Point $O$ lies in the interior of $\Delta PQR$. $O-A-P$,$O-B-Q$,$O-C-R$,$\overline{AB} \parallel \overline{PQ}$ and $\overline{BC} \parallel \overline{QR}$.
To prove: $\overline{AC} \parallel \overline{PR}$.
Proof:
In $\Delta OPQ$,since $\overline{AB} \parallel \overline{PQ}$,by the Basic Proportionality Theorem $(BPT)$,we have:
$\frac{OA}{AP} = \frac{OB}{BQ} \quad \dots (1)$
In $\Delta OQR$,since $\overline{BC} \parallel \overline{QR}$,by the Basic Proportionality Theorem $(BPT)$,we have:
$\frac{OB}{BQ} = \frac{OC}{CR} \quad \dots (2)$
From equations $(1)$ and $(2)$,we get:
$\frac{OA}{AP} = \frac{OC}{CR}$
In $\Delta OPR$,since $\frac{OA}{AP} = \frac{OC}{CR}$,by the converse of the Basic Proportionality Theorem $(BPT)$,we have:
$\overline{AC} \parallel \overline{PR}$.

(N/A) In $\triangle ABC$,$AE$ is the angle bisector of $\angle BAC$. By the Angle Bisector Theorem,we have $\frac{BE}{EC} = \frac{AB}{AC}$.
In $\triangle ADC$,$AF$ is the angle bisector of $\angle DAC$. By the Angle Bisector Theorem,we have $\frac{DF}{FC} = \frac{AD}{AC}$.
Since it is given that $AB = AD$,we can substitute $AD$ with $AB$ in the second equation: $\frac{DF}{FC} = \frac{AB}{AC}$.
Comparing the two equations,we get $\frac{BE}{EC} = \frac{DF}{FC}$.
By the Converse of the Basic Proportionality Theorem (Thales's Theorem) in $\triangle BCD$,since $\frac{BE}{EC} = \frac{DF}{FC}$,it follows that $\overline{EF} \parallel \overline{BD}$.

(A) In $\Delta AOB$,$OD$ is the angle bisector of $\angle AOB$. According to the Angle Bisector Theorem,the ratio of the segments of the side opposite to the angle is equal to the ratio of the other two sides of the triangle.
Therefore,$\frac{AD}{DB} = \frac{OA}{OB}$.
Similarly,in $\Delta BOC$,$OE$ is the angle bisector of $\angle BOC$. Thus,$\frac{BE}{EC} = \frac{OB}{OC}$.
In $\Delta COA$,$OF$ is the angle bisector of $\angle COA$. Thus,$\frac{CF}{FA} = \frac{OC}{OA}$.
Multiplying these three equations together:
$\frac{AD}{DB} \times \frac{BE}{EC} \times \frac{CF}{FA} = \frac{OA}{OB} \times \frac{OB}{OC} \times \frac{OC}{OA}$.
$\frac{AD \times BE \times CF}{DB \times EC \times FA} = 1$.
Therefore,$AD \times BE \times CF = DB \times EC \times FA$.

(N/A) In $\Delta ABD$,$DE$ is the bisector of $\angle ADB$. By the Angle Bisector Theorem,$\frac{AE}{EB} = \frac{AD}{DB}$.
Since $\overline{AD}$ is a median,$D$ is the midpoint of $\overline{BC}$,so $DB = DC$. Thus,$\frac{AE}{EB} = \frac{AD}{DC}$.
In $\Delta ADC$,$DF$ is the bisector of $\angle ADC$. By the Angle Bisector Theorem,$\frac{AF}{FC} = \frac{AD}{DC}$.
Comparing the two equations,we get $\frac{AE}{EB} = \frac{AF}{FC}$.
By the Converse of the Basic Proportionality Theorem (Thales Theorem) in $\Delta ABC$,since $\frac{AE}{EB} = \frac{AF}{FC}$,it follows that $\overline{EF} \parallel \overline{BC}$.

(N/A) Let $\triangle ABC$ be a triangle where $AD$ is the angle bisector of $\angle A$ such that $D$ lies on $BC$ and $BD = DC$.
In $\triangle ABD$ and $\triangle ACD$:
$1$. $\angle BAD = \angle CAD$ (Since $AD$ is the angle bisector of $\angle A$).
$2$. $AD = AD$ (Common side).
$3$. $BD = DC$ (Given).
By the Side-Angle-Side $(SAS)$ congruence criterion,$\triangle ABD \cong \triangle ACD$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Therefore,$AB = AC$.
Since two sides of the triangle are equal,$\triangle ABC$ is an isosceles triangle.

(N/A) Given: $\overline{BD}$ is the bisector of $\angle B$,so $\angle EBD = \angle DBC = \frac{1}{2} \angle B$.
$\overline{CE}$ is the bisector of $\angle C$,so $\angle DCE = \angle ECB = \frac{1}{2} \angle C$.
Since $\overline{DE} \parallel \overline{BC}$,by alternate interior angles,$\angle EDB = \angle DBC$.
Thus,$\angle EDB = \angle EBD$,which implies $\Delta EBD$ is isosceles with $EB = ED$.
Similarly,$\angle DEC = \angle ECB$,so $\angle DEC = \angle DCE$,which implies $\Delta DCE$ is isosceles with $DC = ED$.
Therefore,$EB = DC$.
In $\Delta ABC$,since $\overline{DE} \parallel \overline{BC}$,by the Basic Proportionality Theorem (Thales Theorem),$\frac{AE}{EB} = \frac{AD}{DC}$.
Since $EB = DC$,we have $AE = AD$.
Adding $EB$ to both sides: $AE + EB = AD + DC$,which gives $AB = AC$.
Since two sides are equal,$\Delta ABC$ is an isosceles triangle.

(N/A) $1$. In $\Delta ABC$,let $BO$ be the bisector of $\angle B$ and $CO$ be the bisector of $\angle C$. Since $O$ is the intersection of these bisectors,$AO$ is the bisector of $\angle A$ by the property of the incenter of a triangle.
$2$. According to the Angle Bisector Theorem,in $\Delta ABC$,if $AP$ is the bisector of $\angle A$ intersecting $BC$ at $P$,then the ratio of the sides forming the angle is equal to the ratio of the segments of the opposite side.
$3$. Therefore,$\frac{AB}{AC} = \frac{BP}{PC}$.
$4$. This confirms the Angle Bisector Theorem for $\Delta ABC$ where $AP$ is the internal angle bisector of $\angle A$.

(N/A) Let $ABCD$ be a convex quadrilateral. Let $P, Q, R,$ and $S$ be the midpoints of sides $AB, BC, CD,$ and $DA$ respectively.
Join $AC$. In $\triangle ABC$,$P$ and $Q$ are the midpoints of $AB$ and $BC$. By the Midpoint Theorem,$PQ \parallel AC$ and $PQ = \frac{1}{2} AC$ (Equation $1$).
In $\triangle ADC$,$S$ and $R$ are the midpoints of $AD$ and $CD$. By the Midpoint Theorem,$SR \parallel AC$ and $SR = \frac{1}{2} AC$ (Equation $2$).
From Equation $1$ and Equation $2$,we get $PQ \parallel SR$ and $PQ = SR$.
Since one pair of opposite sides of quadrilateral $PQRS$ is parallel and equal,$PQRS$ is a parallelogram.

(N/A) $1$. In $\triangle ABD$,we are given $\frac{AP}{PB} = \frac{AS}{SD}$. By the Converse of the Basic Proportionality Theorem (Thales Theorem),$PS \parallel BD$.
$2$. In $\triangle BCD$,we are given $\frac{CB}{QB} = \frac{CR}{RD}$. This can be rewritten as $\frac{BQ}{QC} = \frac{DR}{RC}$. By the Converse of the Basic Proportionality Theorem,$QR \parallel BD$.
$3$. Since $PS \parallel BD$ and $QR \parallel BD$,by the property that lines parallel to the same line are parallel to each other,we conclude that $PS \parallel QR$.

(N/A) $1$. In $\Delta ADB$,$DP$ is the bisector of $\angle ADB$. By the Angle Bisector Theorem,$\frac{AP}{PB} = \frac{AD}{BD}$,which implies $AP = \frac{AD \times PB}{BD}$.
$2$. In $\Delta ADC$,$DQ$ is the bisector of $\angle ADC$. By the Angle Bisector Theorem,$\frac{AQ}{QC} = \frac{AD}{DC}$,which implies $AQ = \frac{AD \times QC}{DC}$.
$3$. Multiplying these two equations: $AP \times AQ = \frac{AD \times PB}{BD} \times \frac{AD \times QC}{DC} = \frac{AD^2 \times PB \times QC}{BD \times DC}$.
$4$. Rearranging gives $AP \times AQ \times BD \times DC = AD^2 \times PB \times QC$.
$5$. If $\overleftrightarrow{PQ} \parallel \overleftrightarrow{BC}$,then by the Basic Proportionality Theorem in $\Delta ABC$,$\frac{AP}{PB} = \frac{AQ}{QC}$.
$6$. From the Angle Bisector Theorem,$\frac{AP}{PB} = \frac{AD}{BD}$ and $\frac{AQ}{QC} = \frac{AD}{DC}$.
$7$. Therefore,$\frac{AD}{BD} = \frac{AD}{DC}$,which implies $BD = DC$. Thus,$D$ is the midpoint of $\overline{BC}$.

(A) Let the vertices of the triangle be $A(x_1, y_1)$,$B(x_2, y_2)$,and $C(x_3, y_3)$.
The centroid $G$ is given by $G = (\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3})$.
Given $3 AN = 2 AC$,we have $\frac{AN}{AC} = \frac{2}{3}$,which implies $N$ divides $AC$ in the ratio $2:1$.
Using the section formula,the coordinates of $N$ are $(\frac{2x_3+x_1}{3}, \frac{2y_3+y_1}{3})$.
Since the line $MN$ passes through $G$,the points $M, G, N$ are collinear.
Using the property of the centroid and the ratio of segments,by applying Menelaus' theorem or vector geometry on $\Delta ABC$ with transversal $MGN$,we find the ratio $AM:MB$.
Given the condition $3 AN = 2 AC$,it implies $N$ is such that $AN = \frac{2}{3} AC$.
By applying the property of transversals through the centroid,we obtain $\frac{AM}{MB} = 2$,which means $AM = 2 MB$.

(A) Given that $\square ABCD$ is a trapezium with $\overline{AB} \parallel \overline{CD}$.
We are given $AM \times NC = BN \times MD$,which can be rewritten as $\frac{AM}{MD} = \frac{BN}{NC}$.
Since $\frac{AM}{MD} = \frac{2}{3}$,it follows that $\frac{BN}{NC} = \frac{2}{3}$.
By the Basic Proportionality Theorem (Thales Theorem) or its converse,since $\frac{AM}{MD} = \frac{BN}{NC}$,the line segment $\overline{MN}$ must be parallel to the bases $\overline{AB}$ and $\overline{CD}$.
Let $\overline{MN} \parallel \overline{AB} \parallel \overline{CD}$.
In $\triangle ADC$,since $\overline{MO} \parallel \overline{CD}$,by the intercept theorem (or similarity of triangles $\triangle AMO \sim \triangle ADC$),we have $\frac{AO}{AC} = \frac{AM}{AD}$.
Given $\frac{AM}{MD} = \frac{2}{3}$,let $AM = 2k$ and $MD = 3k$. Then $AD = AM + MD = 2k + 3k = 5k$.
Therefore,$\frac{AO}{AC} = \frac{AM}{AD} = \frac{2k}{5k} = \frac{2}{5}$.

(C) Given that in $\Delta ABC$ and $\Delta PQR$,$\angle A = \angle P$ and $\angle B = \angle R$.
By the $AA$ (Angle-Angle) similarity criterion,$\Delta ABC \sim \Delta PRQ$.
Since the triangles are similar,the ratios of their corresponding sides are equal:
$\frac{AB}{PR} = \frac{AC}{PQ}$
Substitute the given values: $AB = 8$,$PQ = 7.5$,$AC = 6$.
$\frac{8}{PR} = \frac{6}{7.5}$
$PR = \frac{8 \times 7.5}{6}$
$PR = \frac{60}{6} = 10$.
Therefore,$PR = 10$.

(D) In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$.
Since $\overline{MN} \parallel \overline{BC}$,$\angle AMN \cong \angle ABC$ and $\angle ANM \cong \angle ACB$ (Corresponding angles).
By $AA$ similarity criterion,$\Delta AMN \sim \Delta ABC$.
Therefore,the ratio of corresponding sides is equal:
$\frac{AM}{AB} = \frac{MN}{BC}$
We know that $AB = AM + MB = 2 + 3 = 5$.
Substituting the values:
$\frac{2}{5} = \frac{3}{BC}$
$2 \times BC = 3 \times 5$
$2 \times BC = 15$
$BC = \frac{15}{2} = 7.5$

(A) Given that in $\Delta ABC$ and $\Delta XYZ$,$\frac{AB}{XY} = \frac{BC}{YZ}$ and $\angle B \cong \angle Y$.
By the $SAS$ (Side-Angle-Side) similarity criterion,$\Delta ABC \sim \Delta XYZ$.
Since the triangles are similar,the ratios of their corresponding sides are equal:
$\frac{AB}{XY} = \frac{AC}{XZ} = \frac{BC}{YZ}$.
Substituting the given values:
$\frac{6}{15} = \frac{15}{XZ}$.
Cross-multiplying to solve for $XZ$:
$6 \times XZ = 15 \times 15$.
$6 \times XZ = 225$.
$XZ = \frac{225}{6} = 37.5$.
Therefore,$XZ = 37.5$.

(C) To determine the similarity,we compare the ratios of the corresponding sides of $\Delta ABC$ and $\Delta PQR$.
First,arrange the sides of each triangle in increasing order:
For $\Delta ABC$: $AB = 3$,$BC = 6$,$AC = 8$.
For $\Delta PQR$: $PR = 5.1$,$QR = 10.2$,$PQ = 13.6$.
Now,calculate the ratios of the corresponding sides:
Ratio of shortest sides: $\frac{AB}{PR} = \frac{3}{5.1} = \frac{30}{51} = \frac{10}{17}$.
Ratio of medium sides: $\frac{BC}{QR} = \frac{6}{10.2} = \frac{60}{102} = \frac{10}{17}$.
Ratio of longest sides: $\frac{AC}{PQ} = \frac{8}{13.6} = \frac{80}{136} = \frac{10}{17}$.
Since $\frac{AB}{PR} = \frac{BC}{QR} = \frac{AC}{PQ} = \frac{10}{17}$,by the $SSS$ similarity criterion,the correspondence $ABC \leftrightarrow PRQ$ is a similarity.

(C) Given that $\Delta PQR \sim \Delta XYZ$ for the correspondence $PQR \leftrightarrow XZY$.
According to the theorem of areas of similar triangles,the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Therefore,$\frac{\text{Area of } \Delta PQR}{\text{Area of } \Delta XYZ} = \frac{PQ^2}{XZ^2}$.
Substituting the given values: $\frac{24}{\text{Area of } \Delta XYZ} = \frac{8^2}{12^2}$.
$\frac{24}{\text{Area of } \Delta XYZ} = \frac{64}{144}$.
Simplifying the fraction: $\frac{64}{144} = \frac{4}{9}$.
So,$\frac{24}{\text{Area of } \Delta XYZ} = \frac{4}{9}$.
$\text{Area of } \Delta XYZ = \frac{24 \times 9}{4} = 6 \times 9 = 54$.

(N/A) Given: $\Delta ABC \sim \Delta PQR$ with correspondence $ABC \leftrightarrow PQR.$ $\overline{AD}$ and $\overline{PM}$ are medians of $\Delta ABC$ and $\Delta PQR$ respectively.
To prove: $\frac{AD}{PM} = \frac{AB}{PQ}.$
Proof:
$1$. Since $\Delta ABC \sim \Delta PQR,$ the corresponding sides are proportional and corresponding angles are congruent.
$\therefore \frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}$ and $\angle B \cong \angle Q \quad \dots (1)$
$2$. Since $\overline{AD}$ and $\overline{PM}$ are medians,$D$ is the midpoint of $BC$ and $M$ is the midpoint of $QR.$
Therefore,$BC = 2BD$ and $QR = 2QM.$
$3$. Substituting these into equation $(1)$:
$\frac{AB}{PQ} = \frac{2BD}{2QM} = \frac{BD}{QM} \quad \dots (2)$
$4$. In $\Delta ABD$ and $\Delta PQM$:
From $(1)$,$\angle B \cong \angle Q.$
From $(2)$,$\frac{AB}{PQ} = \frac{BD}{QM}.$
$5$. By the $SAS$ (Side-Angle-Side) similarity criterion,$\Delta ABD \sim \Delta PQM.$
$6$. Since the triangles are similar,their corresponding sides are proportional:
$\therefore \frac{AD}{PM} = \frac{AB}{PQ}.$ Hence proved.

(N/A) Given: $\Delta ABC \sim \Delta XYZ$ with correspondence $ABC \leftrightarrow XYZ$. $\overline{AD} \perp \overline{BC}$ and $\overline{XM} \perp \overline{YZ}$.
To prove: $\frac{AD}{XM} = \frac{BC}{YZ}$.
Proof:
$1$. Since $\Delta ABC \sim \Delta XYZ$,the ratios of their corresponding sides are equal and corresponding angles are congruent.
$\therefore \frac{AB}{XY} = \frac{BC}{YZ} = \frac{AC}{XZ}$ and $\angle B \cong \angle Y$ ... $(1)$
$2$. In $\Delta ABD$ and $\Delta XYM$:
$\angle B \cong \angle Y$ (from result $1$)
$\angle ADB \cong \angle XMY$ (both are $90^{\circ}$ as $\overline{AD} \perp \overline{BC}$ and $\overline{XM} \perp \overline{YZ}$)
$3$. By the $AA$ similarity criterion,$\Delta ABD \sim \Delta XYM$.
$4$. Since the triangles are similar,the ratios of their corresponding sides are equal:
$\therefore \frac{AD}{XM} = \frac{AB}{XY}$ ... $(2)$
$5$. From $(1)$ and $(2)$,since $\frac{AB}{XY} = \frac{BC}{YZ}$,we get:
$\frac{AD}{XM} = \frac{BC}{YZ}$.
Hence proved.

(A) In $\Delta ABC$,$m\angle B = 90^\circ$ and $\overline{BM} \perp \overline{AC}$.
$(i)$ For the correspondence $AMB \leftrightarrow ABC$:
$\angle AMB \cong \angle ABC$ (both are $90^\circ$)
$\angle MAB \cong \angle BAC$ (common angle)
Therefore,by the $AA$ similarity criterion,$\Delta AMB \sim \Delta ABC$.
(ii) For the correspondence $BMC \leftrightarrow ABC$:
$\angle BMC \cong \angle ABC$ (both are $90^\circ$)
$\angle MCB \cong \angle BCA$ (common angle)
Therefore,by the $AA$ similarity criterion,$\Delta BMC \sim \Delta ABC$.
(iii) Since $\Delta AMB \sim \Delta ABC$ and $\Delta BMC \sim \Delta ABC$,by the transitivity of similarity,$\Delta AMB \sim \Delta BMC$.