C is the centroid of isosceles triangle ABD with vertex angle ∠ABD. Does the following proof correctly justify that triangles ABE and DBE are congruent? It is given that triangle ABD is isosceles, so segment AB is congruent to DB by the definition of isosceles triangle.

Triangles ABE and DBE share side BE, so it is congruent to itself by the reflexive property.

It is given that C is the centroid of triangle ABD, so segment BE is a perpendicular bisector.

E is a midpoint, creating congruent segments AE and DE, by the definition of midpoint.

Triangles ABE and DBE are congruent by the SSS Postulate.

Triangle ABD with segments BC, DC, and AC drawn from each vertex and meeting at point C inside triangle ABD, segment BC is extended past C with dashed lines so that it intersects with side AD at point E.

There is an error in line 1; segments AB and BC are congruent.

There is an error in line 2; segment BE is not a shared side.

There is an error in line 3; segment BE should be a median.

The proof is correct.

Can someone help me?

x=12 units bd= 10 units

step-by-step explanation:

true

explanation:

the answer will be

9) i can't see the angle so whatever it is just do this: cos(angle) x hypotenuse and then use that answer and they hypotenuse and plug it into this: (answer you just got)^2 + b^2 = (hypotenuse)^2. the first answer will be the answer of the adjacent side and the second answer after this equation will be the opposite side.

10) y = 14 x = √147 or 12.12435565

11) x = 4.8

12) x = 4.2

13) x = 9.3

14) pq: 13.2

qr : 17.6

angle p: 53º

15) angle c: 54.3º

angle a: 35.7º

16) h = 21040

v = 1841

step-by-step explanation:

10) 7/sin(30) = 14 = y

7^2 + b^2 = 14^2

49 + b^2 = 196

b^2 = 196 - 49

√b^2 = √147

x = √147

11) cos(70) x 14 = 4.788282007

x = 4.8

12) tan(25) x 9 = 4.196768923

x = 4.2

13) sin(31) x 18 = 9.270685348

x = 9.3

14) pq: sin(37) x 22 = 13.23993051

pq: 13.2

qr: cos(37) x 22 = 17.56998122

qr : 17.6

angle p: 180 - (37 + 90) = 53º

15) cos-1(7/12) = 54.31466529

angle c: 54.3º

angle a: 180 - (90 + 54.31466529) = 35.68533471

angle a: 35.7º

16) 4 x 5280 = 21120

cos(5) x 21120

h = 21039.63202

h = 21040

sin(5) x 21120

v = 1840.729287

v = 1841