+15 Hi, I'm need help because I don't do well with sentence problems (I don't know what to call them) but I've been struggling for a while, so if you can assist then thanks.
5. Triangle ABC is a right triangle with CD (perpendicular sign) AB. Angle C is a right triangle.
Use the drop down menus to complete the proof of the pythagorean theorem.
By the Choose...Side-Angle-Side Similarity Postulate, Angle-Angle Similarity Postulate, or Side-Side-Side Similarity Postulate, △ACB∼△ADC and △ACB∼△CDB. Since similar triangles have Choose...congruent or proportional sides, BCBA=BDBC and ACAB=ADAC . Using cross multiplication gives the equations (BC)2=(BD)(BA) and (AC)2=(AD)(AB). Adding these together gives(BC)2+(AC)2=(BD)(BA)+(AD)(AB). Factoring out the common segment gives(BC)2+(AC)2=(AB)(BD+AD). Using Choose...segment addition postulate or CPCTC gives (BC)2+(AC)2=(AB)(AB), which simplifies to (BC)2+(AC)2=(AB)2 .
6. Given triangle ACE, Line BD is parallel to AE. Prove BA/CB=DE/CD. Drag an expression or phrase to each box to complete the proof.
StatementReason
△ACE,BD¯¯¯¯¯∥AE¯¯¯¯¯ Given
Corresponding Angles Postulate
△ACE∼△BCD
CA/CB=CE/CD Definition of similar triangles
CA=CB+BA CE=CD+DE Segment Addition Postulate
CB+BA/CB=CD+DE/CD Substitution Property of Equality
CBCB+BACB=CDCD+DECD Addition of fractions
1+BA/CB=1+DE/CD Simplification of fractions
Subtraction Property of Equality
Every empty space is supposed to be filled in. So, the ones that aren't are substraction property of equality, triangle ACE~ triangle BCD, and corresponding angles postulate.
Here are the choices to choose from for the three options (one for each).
-AA similarity postulate
-SAS similarity postulate
-BA/CB=DE/CD
-<4~/=<1, <3~/=<2
-<4~/=<2,<3~/=<1
