Exercises from chapter 12 — Thales' Theorem

Determine the value of $x$ in each case below, with $r$, $s$ and $t$ parallel lines.

In the figures, the lines $r$, $s$ and $t$ are parallel. Determine the values ​​of $x$ and $y$.

In the figure, $\overline{MN}$ is parallel to the base $\overline{BC}$ of the triangle $ABC$. Calculate the value of $x$.

In the figure, $\overline{MN} \parallel \overline{BC}$. Calculate the value of $AB$.

In the figure, calculate the value of $x$.

In the figure, the segments $\overline{AB}$, $\overline{BC}$, $\overline{CD}$ and $\overline{DE}$ measure, respectively, 8 cm, 10 cm, 12 cm and 15 cm. Calculate the measurements of the segments $\overline{A^{\prime}B^{\prime}}$, $\overline{B^{\prime}C^{\prime}}$, $\overline{C^{\prime}D^{\prime}}$ and $\overline{D^{\prime}E^{\prime}}$, knowing that $\overline{A^{\prime}E^{\prime}}$ measures 54 cm and that the straight lines $a$, $b$, $c$, $d$, $e$ are parallel.

In the figure, $r \parallel s \parallel t$. Determine the measurements $x$ and $y$, knowing that they are proportional to 2 and 3, that the segment $\overline{A^{\prime}C^{\prime}}$ measures 30 cm and that the straight lines $a$ and $b$ are parallel.

A beam of 4 parallels determines on a transversal three segments measuring 5 cm, 6 cm and 9 cm. Determine the lengths of the segments that this same beam determines over another transversal, knowing that the segment between the first and fourth parallel measures 60 cm.

See the resolution of the book

$$\frac{x}{5} = \frac{60}{20} \Rightarrow x = 15 \text{ cm}; \qquad \frac{y}{6} = \frac{60}{20} \Rightarrow y = 18 \text{ cm}; \qquad \frac{z}{9} = \frac{60}{20} \Rightarrow z = 27 \text{ cm}.$$

A beam of five parallels determines on a transversal four segments measuring 5 cm, 8 cm, 11 cm and 16 cm. Calculate the length of the segments that this same beam determines over another transversal, knowing that the segment between the extreme parallels measures 60 cm.

A triangle $ABC$ has sides $\overline{AC}$ and $\overline{BC}$ measuring 24 cm and 20 cm. On the side $\overline{AC}$, 6 cm from the vertex $C$, we take a point $M$. Determine the distance from a point $N$ from $\overline{BC}$ to the vertex $C$, so that $\overline{MN}$ is parallel to $\overline{AB}$.

In the triangle $ABC$, the side $\overline{AC}$ measures 32 cm and the side $\overline{BC}$, 36 cm. Through a point $M$ of $\overline{AC}$, 10 cm from the vertex $C$, we draw the parallel to the side $\overline{AB}$, which divides $\overline{BC}$ into two segments $\overline{BN}$ and $\overline{CN}$. Determine the measurement of $\overline{CN}$.

In the figure below, where $a \parallel b \parallel c \parallel d$, we have $AD + AG + HK + KN = 180$ cm, $\dfrac{AE}{AB} = \dfrac{3}{2}$, $\dfrac{JK}{AB} = \dfrac{9}{5}$ and $\dfrac{KL}{AB} = \dfrac{27}{10}$; $\overline{AB}$, $\overline{BC}$ and $\overline{CD}$ are proportional to 2, 3 and 4. Calculate the measurements of the segments $\overline{EF}$, $\overline{LM}$ and $\overline{CD}$.

Three lots face street “A” and street “B”, as in the figure. The side boundaries are perpendicular to “A” street. What is the frontage to street “B” of each lot, knowing that the total frontage to this street is 180 m?

Determine $x$ and $y$, with $r$, $s$ and $t$ parallel lines.

Given a triangle $ABC$ and a segment $\overline{DE}$, with $D$ in $\overline{AB}$ and $E$ in $\overline{AC}$, prove that if $AD : DB = AE : EC$, then $\overline{DE}$ is parallel to $\overline{BC}$.

If $\overline{AS}$ is angle bisector of $\hat{A}$, calculate $x$ in the cases:

If $\overline{AP}$ is angle bisector of the exterior angle at $A$, determine $x$.

In the figure, $\overline{AS}$ is angle bisector inside the angle $\hat{A}$. Calculate the value of $x$.

In the figure, $\overline{AS}$ is angle bisector inside the angle $\hat{A}$. Calculate $x$.

In the figure, calculate the values ​​of $x$ and $y$, with $\overline{BS}$ being the internal angle bisector of the angle $\hat{B}$.

In the figure, $\overline{AD}$ is angle bisector outside the angle $\hat{A}$. Calculate $x$.

Determine the measurement of the side $\overline{AB}$ of the triangle $ABC$: a) $\overline{AS}$ is angle bisector and the perimeter of $\triangle ABC$ is 75 cm; b) $\overline{AP}$ is angle bisector of the external angle in $A$ and the perimeter of $\triangle ABC$ is 23 m.

In the triangle $ABC$ in the figure, $\overline{AS}$ is angle bisector inside the angle $\hat{A}$ and $\overline{AP}$ is angle bisector outside. Calculate the measurement of the segment $\overline{SP}$.

See the resolution of the book

$$\text{bissetriz interna: } \frac{x}{20} = \frac{30 - x}{40} \Rightarrow x = 10; \qquad \text{bissetriz externa: } \frac{y}{20} = \frac{30 + y}{40} \Rightarrow y = 30.$$

Logo $SP = x + y = 40$.

The sides of a triangle measure 5 cm, 6 cm and 7 cm. By how much does the shorter side need to be extended so that it meets the angle bisector of the opposite exterior angle?

Being $\overline{AS}$ and $\overline{AP}$ angle bisectors the internal and external angles in $A$, determine the value of $CP$, given $BS = 8$ m and $SC = 6$ m.

The internal angle bisector of angle $\hat{A}$ of a triangle $ABC$ divides the opposite side into two segments measuring 9 cm and 16 cm. Knowing that $\overline{AB}$ measures 18 cm, determine the measurement of $\overline{AC}$.

The perimeter of a triangle $ABC$ is 100 m. The internal angle bisector of angle $\hat{A}$ divides the opposite side $\overline{BC}$ into two segments of 16 m and 24 m. Determine the sides of this triangle.

The internal angle bisector $\overline{AD}$ of a triangle $ABC$ divides the opposite side into two segments $\overline{BD}$ and $\overline{CD}$ measuring 24 cm and 30 cm. With $AB$ and $AC$ equal to $2x + 6$ and $3x$, respectively, determine the value of $x$ and the measurements of $\overline{AB}$ and $\overline{AC}$.

The external angle bisector $\overline{AS}$ of a triangle $ABC$ determines on the extension of the side $\overline{BC}$ a segment $\overline{CS}$ of measurement $y$. Since the sides $\overline{AB}$ and $\overline{AC}$, respectively, are triple and double the smallest segment determined by angle bisector internal $\overline{AP}$ on the side $\overline{BC}$, which measures 20 cm, determine the value of $y$.

The sides of a triangle measure 8 cm, 10 cm and 12 cm. By how much do you need to extend the shortest side so that it meets the angle bisector of the external angle opposite that side?

Considering the measurements indicated in the figure and knowing that the circle is inscribed in the triangle, determine $x$.

Let us consider a triangle $ABC$ with 15 cm in perimeter. The external angle bisector of the angle $\hat{A}$ of this triangle meets the extension of the side $\overline{BC}$ at a point $S$. Knowing that the internal angle bisector of the angle $\hat{A}$ determines on $\overline{BC}$ two segments $\overline{BP}$ and $\overline{PC}$ measuring 3 cm and 2 cm, respectively, determine the measurements of the sides of the triangle and the measurement of the segment $\overline{CS}$.