Triangle Similarity Theorems and Pythagorean Theorem
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Geometry › Triangle Similarity Theorems and Pythagorean Theorem
In the diagram, $\triangle ABC$ is a right triangle with right angle at $C$. Segment $CD$ is drawn from $C$ to the hypotenuse $AB$, and $CD \perp AB$ at $D$. The legs are labeled $AC=b$ and $BC=a$, and the hypotenuse is labeled $AB=c$. Using triangle similarity created by the altitude (not memorization of a formula), which conclusion can be proven using the diagram?
$a+b=c$
$CD=\dfrac{a+b}{2}$
$AC \cong BC$
$a^2+b^2=c^2$
Explanation
The skill is using triangle similarity to prove the Pythagorean Theorem in a right triangle with an altitude to the hypotenuse. In right triangle ABC with right angle at C, the altitude from C to hypotenuse AB at D creates three similar triangles: ABC, ACD, and BCD. The corresponding sides are proportional, such as the hypotenuse of the large triangle corresponding to the adjacent leg in the smaller triangles. From similarity of ABC and BCD, we set up BC/AB = DB/BC, or a/c = y/a. Adding the relations a² = c y and b² = c x with x + y = c derives a² + b² = c². A common misconception is that the legs are congruent, leading to AC ≅ BC, but this only holds if the right triangle is isosceles. To transfer this strategy, focus on identifying similar triangles by shared angles and setting up proportions rather than memorizing formulas.
In the diagram, $\triangle ABC$ is right at $C$, and altitude $CD$ is drawn to hypotenuse $AB$ with $CD \perp AB$. Let $AD=x$, $DB=y$, and $AB=c$. Which proportion leads to the Pythagorean relationship by combining results from the similar triangles?
$\dfrac{AC}{BC}=\dfrac{AB}{AD}$
$\dfrac{AB}{BC}=\dfrac{BC}{DB}$
$\dfrac{AD}{CD}=\dfrac{CD}{AB}$
$\dfrac{AB}{AC}=\dfrac{DB}{AC}$
Explanation
The skill is using triangle similarity to prove the Pythagorean Theorem in a right triangle with an altitude to the hypotenuse. In right triangle ABC with right angle at C, the altitude from C to hypotenuse AB at D creates three similar triangles: ABC, ACD, and BCD. The corresponding sides are proportional, such as the hypotenuse to the leg adjacent to the shared angle. From similarity of ABC and BCD, we set up AB/BC = BC/DB, or c/a = a/y. This gives a² = c y, which combines with b² = c x and x + y = c to derive a² + b² = c². A common misconception is setting AB/AC = DB/AC, which mismatches corresponding sides. To transfer this strategy, focus on identifying similar triangles by shared angles and setting up proportions rather than memorizing formulas.
A right triangle $\triangle ABC$ is drawn with a right-angle mark at $A$. The altitude $AD$ is drawn from $A$ to the hypotenuse $BC$ and is marked perpendicular at $D$. The legs are labeled $AB=b$ and $AC=a$, and the hypotenuse is labeled $BC=c$. The diagram is not drawn to scale.
Which relationship between side lengths must be true based on the similarity created by the altitude?
$AD^2=BD+DC$
$a^2=c\cdot BD$
$b^2=c\cdot BD$
$BD=DC$
Explanation
The skill here is proving the Pythagorean Theorem using similarity in right triangles with an altitude to the hypotenuse. When the altitude AD is drawn from the right angle at A to the hypotenuse BC in right triangle ABC, it creates three similar triangles: ABC, ABD, and ADC. Corresponding sides are identified by matching angles, such as the shared angle at B in triangles ABC and ABD, and the shared angle at C in triangles ABC and ADC. This similarity sets up proportions like AB/BC = BD/AB and AC/BC = DC/AC. Multiplying both sides of these proportions derives AB² = BC · BD and AC² = BC · DC, and adding them gives AB² + AC² = BC · (BD + DC) = BC², proving the Pythagorean Theorem. A common distractor misconception is assuming BD = DC in all cases, which only holds for isosceles right triangles. To transfer this strategy, focus on the structural similarities from the altitude rather than memorizing the theorem.
In the diagram, $\triangle ABC$ is right with $\angle A$ marked $90^\circ$. An altitude $AD$ is drawn to the hypotenuse $BC$ and is marked perpendicular at $D$. The hypotenuse is labeled $BC=c$ and is split into $BD=m$ and $DC=n$ (these are labels, not computed values). No other lengths are provided, and the diagram is not drawn to scale.
Which relationship between side lengths must be true from triangle similarity?
$c=m+n$
$AD=m+n$
$m^2+n^2=c^2$
$AD^2=mn$
Explanation
The skill here is proving the Pythagorean Theorem using similarity in right triangles with an altitude to the hypotenuse. When the altitude AD is drawn from the right angle at A to the hypotenuse BC in right triangle ABC, it creates three similar triangles: ABC, ABD, and ADC. Corresponding sides are identified by matching angles, such as the shared angle at B in triangles ABC and ABD, and the shared angle at C in triangles ABC and ADC. This similarity sets up proportions like AB/BC = BD/AB and AC/BC = DC/AC, and also AD/BD = DC/AD for the geometric mean. Multiplying relevant proportions derives AB² = BC · BD, AC² = BC · DC, and AD² = BD · DC, proving key relationships including the Pythagorean Theorem by addition. A common distractor misconception is assuming m² + n² = c², which misapplies the theorem to the segments. To transfer this strategy, focus on the structural similarities from the altitude rather than memorizing the theorem.
In right triangle $\triangle ABC$, the right angle at $A$ is marked. The altitude $AD$ is drawn to hypotenuse $BC$ with a right-angle mark at $D$ indicating $AD\perp BC$. The legs are labeled $AB=b$ and $AC=a$, and the hypotenuse is labeled $BC=c$. The diagram is not drawn to scale.
Which proportion leads to the Pythagorean relationship by adding two similarity-based equations?
$\dfrac{AB}{BC}=\dfrac{BD}{AB}$ and $\dfrac{AC}{BC}=\dfrac{DC}{AC}$
$\dfrac{AB}{AC}=\dfrac{BC}{AD}$ and $\dfrac{AC}{AB}=\dfrac{BC}{AD}$
$\dfrac{AB}{BC}=\dfrac{AC}{AB}$
$\dfrac{BD}{DC}=\dfrac{AB}{AC}$ and $\dfrac{DC}{BD}=\dfrac{AB}{AC}$
Explanation
The skill here is proving the Pythagorean Theorem using similarity in right triangles with an altitude to the hypotenuse. When the altitude AD is drawn from the right angle at A to the hypotenuse BC in right triangle ABC, it creates three similar triangles: ABC, ABD, and ADC. Corresponding sides are identified by matching angles, such as the shared angle at B in triangles ABC and ABD, and the shared angle at C in triangles ABC and ADC. This similarity sets up proportions like AB/BC = BD/AB and AC/BC = DC/AC. Multiplying both sides of these proportions derives AB² = BC · BD and AC² = BC · DC, and adding them gives AB² + AC² = BC · (BD + DC) = BC², proving the Pythagorean Theorem. A common distractor misconception is equating unrelated ratios like AB/BC = AC/AB without basis in similarity. To transfer this strategy, focus on the structural similarities from the altitude rather than memorizing the theorem.
In the diagram, $\triangle ABC$ is a right triangle with $\angle A$ marked as a right angle. Segment $AD$ is drawn from $A$ to the hypotenuse $BC$ and is marked perpendicular to $BC$ at $D$ (so $AD$ is an altitude). The legs are labeled $AB=b$ and $AC=a$, and the hypotenuse is labeled $BC=c$. The hypotenuse is split into $BD$ and $DC$ (no numeric lengths are given). The diagram is not drawn to scale.
Which equation follows from the similarity of the triangles formed by the altitude?
$AB=AC$
$AD=c$
$\dfrac{BD}{DC}=\dfrac{a}{b}$
$a^2+b^2=c^2$
Explanation
The skill here is proving the Pythagorean Theorem using similarity in right triangles with an altitude to the hypotenuse. When the altitude AD is drawn from the right angle at A to the hypotenuse BC in right triangle ABC, it creates three similar triangles: ABC, ABD, and ADC. Corresponding sides are identified by matching angles, such as the shared angle at B in triangles ABC and ABD, and the shared angle at C in triangles ABC and ADC. This similarity sets up proportions like AB/BC = BD/AB and AC/BC = DC/AC. Multiplying both sides of these proportions derives AB² = BC · BD and AC² = BC · DC, and adding them gives AB² + AC² = BC · (BD + DC) = BC², proving the Pythagorean Theorem. A common distractor misconception is assuming BD/DC = a/b directly, but similarity shows BD/DC = b²/a² instead. To transfer this strategy, focus on the structural similarities from the altitude rather than memorizing the theorem.
A right triangle $\triangle ABC$ is shown with the right angle marked at $A$. The altitude $AD$ is drawn to the hypotenuse $BC$ with $AD\perp BC$ marked at $D$. The legs are labeled $AB=b$ and $AC=a$, and the hypotenuse is labeled $BC=c$. The diagram is not drawn to scale.
Which statement justifies the Pythagorean Theorem by using similarity rather than stating the formula?
Use the fact that $AD$ is perpendicular to $BC$ to claim $AB=BD$ and $AC=DC$.
Use $a^2+b^2=c^2$ because it applies to every right triangle.
Use similarity to show $AB^2=BC\cdot BD$ and $AC^2=BC\cdot DC$, then add to get $AB^2+AC^2=BC^2$.
Use congruence of $\triangle ADB$ and $\triangle ADC$ to conclude $AB^2+AC^2=BC^2$.
Explanation
The skill here is proving the Pythagorean Theorem using similarity in right triangles with an altitude to the hypotenuse. When the altitude AD is drawn from the right angle at A to the hypotenuse BC in right triangle ABC, it creates three similar triangles: ABC, ABD, and ADC. Corresponding sides are identified by matching angles, such as the shared angle at B in triangles ABC and ABD, and the shared angle at C in triangles ABC and ADC. This similarity sets up proportions like AB/BC = BD/AB and AC/BC = DC/AC. Multiplying both sides of these proportions derives AB² = BC · BD and AC² = BC · DC, and adding them gives AB² + AC² = BC · (BD + DC) = BC², proving the Pythagorean Theorem. A common distractor misconception is assuming congruence between triangles ADB and ADC instead of similarity. To transfer this strategy, focus on the structural similarities from the altitude rather than memorizing the theorem.