Triangles A B C and A B F are congruent. Triangle A B C is reflected across line B A to form triangle A B F.

Lines e and f are intersected by line a. At the intersection of lines a and e, the bottom right angle is 72 degrees. At the intersection of lines a and f, the uppercase right angle is 108 degrees.

In the diagram, mFLI is 106°, mFLG = (2x – 1)°,mGLH = (x + 17)°, and mHLI = (4x – 15)°.

Law of sines:





Triangle R S T is shown. Angle T R S is a right angle. The length of R T is 5, the length of R S is 12, and the length of hypotenuse S T is 13.





The equation can be used to find the length of .

Given: and g is a transversalProve:

The vertices of triangle ABC are A(5,2), B(–1,–6) and C(1,5). If ΔDEF is similar to ΔABC and , which could be the coordinates of vertices D and E?

Rhombus LMNO is shown with its diagonals.
Figure ABCD is a square. Prove BD ≅ AC. Statements Reasons1.ABCD is a square1.given2.∠DAB, ∠ABC, ∠BCD, and ∠CDA are right angles2.definition of a square3.∠DAB ≅ ∠ABC ≅ ∠BCD ≅ ∠CDA3.right angles are congruent4.AB ≅ BC ≅ CD ≅ DA4.?5.△BAD ≅ △ABC5.SAS6.BD ≅ AC6.CPCTCWhat is the missing reason in the proof?

A transformation of ΔDEF results in ΔD'E'F'.

Which points lie on the line that passes through point P and is parallel to the given line? Select three options.
Which rules could describe the rotation? Select two options.
Which defines a line segment?
Quadrilateral A B C D is shown. Sides A D and B C are parallel. Sides A B and C D are congruent. Angle A is 115 degrees.
Consider the diagram.

Point O is the center of dilation. Triangle D E F is dilated to create smaller triangle D prime E prime F prime. The length of O F prime is 3. The length of F prime F is 7.




Parallelogram P Q S R is shown. The length of P Q is (2 x + 5) centimeters and the length of R S is (4 x + 1) centimeters.

Triangle QRS is to be dilated using the rule .

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