# Questions tagged [congruences-geometry]

For questions about congruent geometric figures, e.g., determining whether one figure is congruent to another.

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### Does there exist an efficient way to determine if two $n$-simplices in $\mathbb{R}^n$ are congruent?

Given two $n$-simplices in $\mathbb{R}^n$, does there exist an algorithm to determine if they are congruent which has a time complexity of the form $O(n^c)$ where $c$ is a constant? I'm assuming that ...
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### Prove ABC,A'B'C' are congruent:D is on BC,D' is on B'C', $\angle BAD \angle CAD= \angle B'A'D' \angle C'A'D', AB=A'B', AC=A'C', AD=A'D'$

In $\triangle ABC$ and $\triangle A'B'C'$, $D$ is a point on line segment $BC$ and $D'$ is a point on line segment $B'C'$. $\frac{\angle BAD}{\angle CAD}=\frac{\angle B'A'D'}{\angle C'A'D'}$, $AB=A'B'$...
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### Triangle and Median related question

Let BD be a median in triangle ABC. The points E and F divide the median BD in three equal parts, such that BE = EF = FD. If AB = 1 and AF = AD, find the length of the line segment CE. I have tried ...
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### Find the measure of $\angle G$ in a triangle

In the following figure, $\Delta$ADB, $\Delta$PCB and $\Delta$EFG are right triangles. $PB=AE$,$AC=CB$. Question: What is the value of $\angle G$? I figured out $\angle A = \angle CPB$ , $EF=AC$. ...
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### Geometry find the angle

Good day, excuse me, I am looking at congruence themes in Geometry, and I am trying to solve this problem, but I am not sure how to proceed. The coloured numbers is what I have so far (although now ...
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### Similarity by SSA (side-side-angle) in obtuse triangles

Source: Challenge and Thrills of Pre-College Mathematics, Page 74, Problem 54: "In two obtuse triangles, an acute angle of the one is equal to an angle of the other, and sides about the other ...
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### Finding side lengths of a trapezoid given the distance between its diagonal intersection and midpoint of a diagonal

The Question My friend recently gave me a problem that I was interested in, but could not completely solve. I've already constructed a diagram that roughly represents the problem, and it's below. ...
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### Let $ABCD$ be a square. Points $A,B,G$ are collinear. $AC$ and $DG$ meet at $E$ , and $DG$ and $BC$ meet at $F$.

Let $ABCD$ be a square. Points $A,B,G$ are collinear. $AC$ and $DG$ meet at $E$ , and $DG$ and $BC$ meet at $F$. If $DE = 15$ cm, $EF = 9$ cm, find $FG$. What I Tried: Here is a picture :- I have ...
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### In the diagram, $AB$ $||$ $EF$ $||$ $DC$ . Given that $AC + BD = 250$ , $BC = 100$ and $EC + ED = 150$, find $CF$.

In the diagram, $AB$ $||$ $EF$ $||$ $DC$ . Given that $AC + BD = 250$, $BC = 100$ and $EC + ED = 150$, find $CF$. What I Tried: Here is the diagram :- I have assigned variables for the different ...
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### $\Delta ABC$, $AC = 2BC$ and $\angle C = 90^\circ$ and $D$ is the foot of the altitude from $C$ onto $AB$. Find $AE : EC$ .

$\Delta ABC$, $AC = 2BC$ and $\angle C = 90^\circ$ and $D$ is the foot of the altitude from $C$ onto $AB$. A circle with diameter $AD$ intersects the segment $AC$ at $E$. Find $AE : EC$. What I Tried:...
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### In $\Delta ABC,$ side $AC$ and the perpendicular bisector of $BC$ meet at $D$, where $BD$ bisects $\angle ABC$.

In $\Delta ABC,$ side $AC$ and the perpendicular bisector of $BC$ meet at $D$, where $BD$ bisects $\angle ABC$. If $CD = 7$ and $[\Delta ABD] = a\sqrt{5}$ , find $a$ . What I Tried: Here is a picture:...
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### In $\Delta ABC$, $AC = BC$ and $\angle C = 120^\circ$. $M$ is on side $AC$ and $N$ is on side $BC$ .

In $\Delta ABC$, $AC = BC$ and $\angle C = 120^\circ$. $M$ is on side $AC$ and $N$ is on side $BC$ such that $\angle BAN = 50^\circ$ and $\angle ABM = 60^\circ$. Find $\angle NMB$ . What I Tried: ...
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### In right $\Delta ABC$, $\angle C = 90^\circ$. $E$ is on $BC$ such that $AC = BE$. $D$ is on $AB$ such that $DE \perp BC$ .

In right $\Delta ABC$, $\angle C = 90^\circ$. $E$ is on $BC$ such that $AC = BE$. $D$ is on $AB$ such that $DE \perp BC$ . Given that $DE + BC = 1$ and $BD = \frac{1}{2}$, find $\angle B$. What I ...
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### In an isosceles right $\Delta ABC$, $\angle B = 90^\circ$. AD is the median on BC. Let $AB = BC = a$.

In an isosceles right $\Delta ABC$, $\angle B = 90^\circ$. AD is the median on BC. Let $AB = BC = a$. If $BE \perp AD$, intersecting $AC$ at $E$, and $EF \perp BC$ at $F$, find $EF$ in terms of $a$. ...
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### $ABCD$ is a rectangle of area $210$ cm$^2$. $L$ is a mid-point of $CD$ . $P,Q$ trisect $AB$ . $AC$ cuts $LP,LQ$ at $M,N$ respectively.

$ABCD$ is a rectangle of area $210$ cm$^2$. $L$ is a mid-point of $CD$ . $P,Q$ trisect $AB$ . $AC$ cuts $LP,LQ$ at $M,N$ respectively. Find $[\Delta LMN]$ What I Tried: For a geometry Problem, it's ...
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### $ABCD$ is a square. $E$ is the midpoint of $CB$, $AF$ is drawn perpendicular to $DE$. If the side of the square is $2016$ cm , find $BF$.

$ABCD$ is a square. $E$ is the midpoint of $CB$, $AF$ is drawn perpendicular to $DE$. If the side of the square is $2016$ cm , find $BF$. What I Tried: Here is a picture, I used a really peculiar ...
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### is still there a contradiction in the proof by contradiction of the proposition 6 of Elements using this triangle?

The proposition 6 says: If in a triangle two angles equal one another, then the sides opposite the equal angles also equal one another. And the proof of Euclid in his Elements basically is (for ...
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### Is the criterion SAA of congruent triangles valid?

I think the fundamental criterions for triangles congruence are: SAS (Side-Angle-Side) ASA (Angle-Side-Angle) SSS (Side-Side-Side) But some proofs like this one: https://www.youtube.com/watch?v=...
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### Prove that ED=EF [closed]

In the diagram below $AD=DC$ and $AE=EB$ and both triangles $AEB$ and $ADC$ are right angel triangles and $M$ is the midpoint of $BC$ also $MD=MF$. Prove that $ED=EF$
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### Find the length x such that the two distances in the triangle are the same

I have been working on the following problem Statement Assume you have a right angle triangle $\Delta ABC$ with cateti $a$, $b$ and hypotenuse $c = \sqrt{a^2 + b^2}$. Find or construct a point $D$ ...
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### If ratio of sides of two triangles is constant then the triangles have the same angles

If $\triangle ABC$ and $\triangle A'B'C'$ are a pair of triangles such that $$\dfrac{|AB|}{|A'B'|}=\dfrac{|BC|}{|B'C'|}=\dfrac{|AC|}{|A'C'|}$$ then $$\triangle ABC \sim \triangle A'B'C'$$ I have ...