If segment PD is parallel to segment CR then triangle PDX is similar to triangle CRX, due to angle PXD and angle CXR being congruent by the vertical angle theorem, and angle PDX and angle CRX being congruent by the alternate interior angle theorem, these 2 triangles are similar by the A.A similarity postulate.
Answer:
D. SAS
Step-by-step explanation:
Given: ΔABC
Bisecting <ABC to create point D implies that BD is a common side to ΔABD and ΔCBD.
Also,
m<ABD = m<CBD (angle bisector)
BA = BC (given property of the isosceles triangle)
Therefore,
ΔABD ≅ ΔCBD (Side Angle Side)
The reason for statement 5 in this proof is that ΔABD ≅ ΔCBD by SAS (Side-Angle-Side) relations of the congruent triangles.
Answer:
Step-by-step explanation:
4y - 2(5 - y + 4) = 4y - 2(9 - y)
= 4y + 9*(-2) - y *(-2)
= 4y - 18 + 2y {Combine like terms 4y and 2y}
= 6y - 18
6y - 18 = 6*y - 6*3
= 6(y - 3)
6y- 18 = 2 *3y - 2*9
= 2(3y -9)
2(3y - 9) and 6(y- 3 ) are equivalent to 4y - 2(5- y +4)
Others are not equivalent
Answer:
×=18+25=9²
Step-by-step explanation:
Hope this helps
