Euler's Eyes
2026-10-11
Exercises from chapter 5 — Parallelism
Exercises and resolutions
Since the line $a$ is parallel to the line $b$, determine $x$ in the cases:
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a) $50^{\circ}$ · b) $60^{\circ}$
If the lines $r$ and $s$ are parallel, determine $x$ in the cases:
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a) $60^{\circ}$ · b) $70^{\circ}$
The lines $r$ and $s$ in the figure are parallel. Determine $x$ and $y$.
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a) $x = 120^{\circ}$, $y = 75^{\circ}$ · b) $x = 20^{\circ}$, $y = 50^{\circ}$
In the figure, being $a \parallel b$, calculate $\alpha + \beta - \gamma$.
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$30^{\circ}$
The sum of the four acute angles formed by two parallel lines cut by a transversal line is equal to $80^{\circ}$. Determine the obtuse angle.
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$160^{\circ}$
Since $a$ is parallel to $b$, calculate $x$.
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$x = 45^{\circ}$
In the figure, being $a \parallel b$, calculate $x$.
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$x = 7^{\circ}12^{\prime}$
In the figure, being $r \parallel s$, calculate $x$ and $y$.
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$x = 20^{\circ}$ e $y = 30^{\circ}$
In the figure we have the angles $\alpha$ and $\beta$ with respectively parallel sides. Being $\alpha = 8x$ and $\beta = 2x + 30^{\circ}$, determine the supplement of $\beta$.
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$140^{\circ}$ ($x = 5$, $\alpha = \beta = 40^{\circ}$).
Calculate the value of $x + y$, being $r \parallel s$ and $t \parallel v$.
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$x + y = 180^{\circ}$
If the lines $r$ and $s$ are parallel, determine $x$, $y$ and $z$ in the cases:
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a) $x = 50^{\circ}$, $y = 60^{\circ}$, $z = 70^{\circ}$ · b) $x = 40^{\circ}$, $y = z = 120^{\circ}$
Determine the value of $x$ in the following cases:
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a) $50^{\circ}$ · b) $40^{\circ}$
Determine $y$ in cases:
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a) $110^{\circ}$ · b) $120^{\circ}$
Determine $x$ in cases:
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a) $60^{\circ}$ · b) $50^{\circ}$
Determine $x$ and $y$ in the following cases:
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a) $x = 30^{\circ}$, $y = 40^{\circ}$ · b) $x = 30^{\circ}$, $y = 30^{\circ}$
Determine the angles of the triangle in the cases:
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a) $40^{\circ}$, $60^{\circ}$ e $80^{\circ}$ · b) $30^{\circ}$, $60^{\circ}$ e $90^{\circ}$
If the triangle $ABC$ is isosceles with base $\overline{BC}$, determine $x$ in the cases:
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a) $40^{\circ}$ · b) $45^{\circ}$ · c) $120^{\circ}$ · d) $105^{\circ}$
Determine $\alpha + \beta + \gamma$ in cases:
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a) $360^{\circ}$ · b) $900^{\circ}$
The triangle $ABC$ is isosceles with base $\overline{BC}$. Determine the value of $x$ in the following cases:
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a) $50^{\circ}$ · b) $36^{\circ}$ · c) $70^{\circ}$
Determine the value of the unknown (segments with “equal marks” are congruent).
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a) $30^{\circ}$ · b) $55^{\circ}$ · c) $80^{\circ}$ · d) $36^{\circ}$ · e) $105^{\circ}$ · f) $25^{\circ}$ · g) $x = 30^{\circ}$, $y = 40^{\circ}$ · h) $x = 15^{\circ}$, $y = 40^{\circ}$
In the figure below, $\overline{ED}$ is parallel to $\overline{BC}$. With $B\hat{A}E$ equal to $80^{\circ}$ and $A\hat{B}C$ equal to $35^{\circ}$, calculate the measurement of $A\hat{E}D$.
See the resolution of the book
Simply extend $\overline{DE}$ until the straight line $DE$ meets $\overline{AB}$ at a point $P$. Note that $x$ is the external angle of the triangle $APE$. Then $\alpha = 35^{\circ}$ (corresponding to $\hat{B}$, as $\overleftrightarrow{DE} \parallel \overleftrightarrow{BC}$) and
$$x = \alpha + 80^{\circ} = 35^{\circ} + 80^{\circ} = 115^{\circ}.$$Determine the value of $x$ and $y$, being $r \parallel s$.
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$x = 10^{\circ}$ e $y = 150^{\circ}$
Calculate the value of $x$, being $r \parallel s$.
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$x = 72^{\circ}$
If $r \parallel s$, calculate $\alpha$.
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$\alpha = 100^{\circ}$
In the figure below, the lines $r$ and $s$ are parallel. Calculate $\alpha$.
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$\alpha = 52^{\circ}$
In the figure, calculate the measure of the angle $\alpha$, being $r \parallel s$.
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$\alpha = 100^{\circ}$
In the figure, $\overline{AB}$ is parallel to $\overline{CD}$. Being $C\hat{D}B = 150^{\circ}$ and $A\hat{B}C = 25^{\circ}$, calculate $C\hat{B}D$.
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$C\hat{B}D = 5^{\circ}$
Determine the value of $x$.
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$x = 60^{\circ}$
Calculate $x$ in the triangle $ABC$ in the figure.
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$x = 15^{\circ}$
The internal angles of a triangle are proportional to 2, 3 and 4, respectively. Determine the measurement of the largest of them.
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$80^{\circ}$
In exercises 160, 161 and 162, calculate the unknown(s) in the triangle $ABC$.
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$x = 110^{\circ}$
Being $\overline{AB} \equiv \overline{AC}$:
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$x = 55^{\circ}$ e $y = 70^{\circ}$
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$x = 70^{\circ}$
In the figure, the triangle $ABC$ is isosceles with base $\overline{BC}$. Calculate the value of $x$.
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$x = 65^{\circ}$
Calculate $x$ and $y$ indicated in the figure below.
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$x = 70^{\circ}$ e $y = 125^{\circ}$
The figure shows an isosceles triangle $ABC$, with base $\overline{BC}$. Being $\overline{BD}$ angle bisector from $A\hat{B}C$ and $\overline{CD}$ angle bisector from $A\hat{C}B$, calculate the value of $x$.
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$x = 130^{\circ}$
The triangle $ACD$ in the figure is isosceles with base $\overline{AD}$. With $12^{\circ}$ being the measure of angle $B\hat{A}D$ and $20^{\circ}$ being the measure of angle $A\hat{B}C$, calculate the measure of angle $A\hat{C}D$.
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$A\hat{C}D = 116^{\circ}$
An exterior angle of the base of a isosceles triangle is the $\dfrac{5}{4}$ of the vertex angle. Calculate the angles of this triangle.
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$120^{\circ}$, $30^{\circ}$ e $30^{\circ}$
In a isosceles triangle $ABC$, the vertex angle $A$ is worth $\dfrac{1}{10}$ of the sum of the external angles in $B$ and $C$. With $\overline{BC}$ being the base of the triangle, determine the angle $\hat{A}$.
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$\hat{A} = 20^{\circ}$
In a triangle $ABC$, the obtuse angle formed by angle bisectors of angles $\hat{B}$ and $\hat{C}$ exceeds the angle $\hat{A}$ in $76^{\circ}$. Determine $\hat{A}$.
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$\hat{A} = 28^{\circ}$
Prove that in the triangle $ABC$ in the figure the relationship $\alpha - \beta = \hat{B} - \hat{C}$ holds, with $\overline{AD}$ angle bisector being the angle $B\hat{A}C$.
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$\alpha$ is external to $\triangle ABD$: $\alpha = \dfrac{\hat{A}}{2} + \hat{B}$. $\beta$ is external to $\triangle ACD$: $\beta = \dfrac{\hat{A}}{2} + \hat{C}$. Subtracting, $\alpha - \beta = \hat{B} - \hat{C}$.
In a triangle $ABC$, the angle formed by angle bisectors of the angles $\hat{B}$ and $\hat{C}$, opposite $\overline{BC}$, is five times the angle $\hat{A}$. Determine the measure of the angle $\hat{A}$.
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$\hat{A} = 20^{\circ}$
In the figure, calculate the value of $x$ as a function of $m$.
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$x = 2m$
In any triangle $ABC$, the angle opposite $\overline{BC}$ formed by the angle bisectors of the internal angles in $B$ and $C$ is equal to the supplement of the half complement of the vertex angle $A$.
See the resolution of the book
In problems whose statement is a proposition, the property is required to be demonstrated. Let $b$ and $c$ be the halves of $\hat{B}$ and $\hat{C}$ and $x = B\hat{D}C$, where $D$ is the meeting of angle bisectors.
$$\triangle DBC:\ x + b + c = 180^{\circ} \Rightarrow x = 180^{\circ} - (b + c)$$$$\triangle ABC:\ 2b + 2c + \hat{A} = 180^{\circ} \Rightarrow b + c = 90^{\circ} - \frac{\hat{A}}{2}$$
Logo $x = 180^{\circ} - \left(90^{\circ} - \dfrac{\hat{A}}{2}\right)$: the supplement to the complement of $\dfrac{\hat{A}}{2}$. (In other words, $x = 90^{\circ} + \dfrac{\hat{A}}{2}$ — test in exercise 165!)
In the figure, calculate the angle $x$, with $\alpha$ being triple of $\beta$ and $\gamma$ being six times of $\beta$.
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$x = 50^{\circ}$
In a triangle $ABC$, the vertex angle $A$ is equal to the eighth part of the obtuse angle formed by the angle bisectors of the angles adjacent to $\overline{BC}$. Determine the measure of the vertex angle $A$.
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$12^{\circ}$
An external vertex angle of a isosceles triangle measures $150^{\circ}$. Determine: a) the angles of the triangle; b) the obtuse angle formed by angle bisectors of the angles at the base of the triangle; c) the angles formed by angle bisector of one of the base angles and angle bisector of the vertex angle.
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a) $30^{\circ}$, $75^{\circ}$ e $75^{\circ}$ · b) $105^{\circ}$ · c) $52^{\circ}30^{\prime}$ e $127^{\circ}30^{\prime}$
Determine the measure of the smallest angle formed by the external angle bisectors relative to the vertices $B$ and $C$ of a triangle $ABC$, knowing that the angle $\hat{A}$ measures $76^{\circ}$.
See the resolution of the book
Let $b$ and $c$ be the halves of the external angles in $B$ and $C$ and $x$ be the requested angle.
- $\hat{B} + \hat{C} + 76^{\circ} = 180^{\circ} \Rightarrow \hat{B} + \hat{C} = 104^{\circ}$.
- $2b + \hat{B} = 180^{\circ}$ and $2c + \hat{C} = 180^{\circ}$; adding, $2(b + c) + 104^{\circ} = 360^{\circ} \Rightarrow b + c = 128^{\circ}$.
- In the triangle formed by the outer angle bisectors and $\overline{BC}$: $x + b + c = 180^{\circ} \Rightarrow x = 52^{\circ}$.
Determine the measures of the three angles of a triangle, knowing that the second is the $\dfrac{3}{2}$ of the first and that the third is the semisum of the first two.
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$48^{\circ}$, $72^{\circ}$ e $60^{\circ}$
The three angles of a triangle are such that the second measures $28^{\circ}$ less than the first and the third $10^{\circ}$ more than the first. Determine the three angles of the triangle.
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$66^{\circ}$, $38^{\circ}$ e $76^{\circ}$
In a isosceles triangle, the apex angle is half of each of the base angles. Determine the three angles of the triangle.
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$36^{\circ}$, $72^{\circ}$ e $72^{\circ}$
Determine the angle formed by the angle bisectors of two internal side angles of two parallel lines intersected by any transversal.
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$90^{\circ}$
In the figure, determine the measure of the angle $\alpha$ as a function of $m$, being $\hat{A} = 3m$, $\hat{B} = 2m$, $\hat{D} = m$ and $B\hat{C}D = \alpha$.
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$\alpha = 6m$
In any triangle $ABC$, the angle opposite $\overline{BC}$ formed by the angle bisectors of the external angles in $B$ and $C$ is equal to the complement of half the vertex angle $A$ of the triangle.
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Do as in exercise 173, now with the external angle bisectors: the angle is $90^{\circ} - \dfrac{\hat{A}}{2}$.
In the figure, with $\overline{AB}$ congruent to $\overline{AC}$ and $\overline{AE}$ congruent to $\overline{AD}$, calculate the measure of the angle $C\hat{D}E$, given $B\hat{A}D = 48^{\circ}$.
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$C\hat{D}E = 24^{\circ}$
Determine the measure of the angle of the vertex $A$ of isosceles triangle $ABC$, knowing that the segments $\overline{BC}$, $\overline{CD}$, $\overline{DE}$, $\overline{EF}$ and $\overline{FA}$ are congruent.
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$\hat{A} = 20^{\circ}$
In the figure, triangle $ABC$ is equilateral and triangle $CDB$ is isosceles. Calculate the value of $2x + y$, being $B\hat{C}D = x$ and $A\hat{B}D = y$.
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$2x + y = 195^{\circ}$
Consider the triangle $ABC$, where $AB = AC = 5$ cm and $BC = 7$ cm. On the side $\overline{BC}$ we take a point $D$ such that $BD = 3$ cm and through the point $D$ we draw $\overline{DE}$ and $\overline{DF}$ respectively parallel to $\overline{AC}$ and $\overline{AB}$, with $E$ in $\overline{AB}$ and $F$ in $\overline{AC}$. Calculate the perimeter of $AEDF$.
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10 cm
From the figure, we know that $AB = AC$, $\hat{A} = 100^{\circ}$ and $AD = BC$. Determine $x = C\hat{B}D$.
See the resolution of the book
- Let us indicate $AB = AC = b$ and $CD = a$; then $BC = AD = a + b$.
- Let us draw $\overline{AP}$ with $AP = b$, so that $B\hat{A}P = 60^{\circ}$: the triangle $APB$ is equilateral with one side $b$.
- In triangles $PAD$ and $ABC$: $PA = AB = b$, $P\hat{A}D = 100^{\circ} - 60^{\circ} = 40^{\circ} = A\hat{B}C$ and $AD = BC$. In the LAL case, they are congruent. Then $PD = AC = b$ and $A\hat{P}D = B\hat{A}C = 100^{\circ}$.
- Of $PD = b = PB$, $\triangle PBD$ is isosceles. In it $\hat{P} = 60^{\circ} + 100^{\circ} = 160^{\circ}$, then $P\hat{B}D = P\hat{D}B = 10^{\circ}$.
- Finally, from $A\hat{B}P = 60^{\circ}$, $D\hat{B}P = 10^{\circ}$ and $C\hat{B}A = 40^{\circ}$, we conclude that $x = C\hat{B}D = 10^{\circ}$.