Problem In the following triangle, point EEE is the midpoint of \overline{AB}
AB
start overline, A, B, end overline, and point DDD is the midpoint of \overline{AC}
AC
start overline, A, C, end overline.

Below is the proof that DE=\dfrac{1}{2}CBDE=
2
1
​
CBD, E, equals, start fraction, 1, divided by, 2, end fraction, C, B. The proof is divided into four parts, where the title of each part indicates its main purpose.
Complete parts B and C of the proof.
Part A: Prove \dfrac{CA}{DA} = 2
DA
CA
​
=2start fraction, C, A, divided by, D, A, end fraction, equals, 2
[Show the steps.]
Part B: Prove \dfrac{BA}{EA} = 2
EA
BA
​
=2start fraction, B, A, divided by, E, A, end fraction, equals, 2
StatementReason
5BE=EABE=EAB, E, equals, E, ADefinition of midpoint
6BA=BE+EABA=BE+EAB, A, equals, B, E, plus, E, ASegment addition postulate
7BA=EA+EABA=EA+EAB, A, equals, E, A, plus, E, ASubstitution (5, 6)
8\dfrac{BA}{EA} = 2
EA
BA
​
=2start fraction, B, A, divided by, E, A, end fraction, equals, 2Divide both sides by

. (7)
Part C: Prove \triangle CAB \sim \triangle DAE△CAB∼△DAEtriangle, C, A, B, \sim, triangle, D, A, E
StatementReason
9\angle DAE \cong \angle∠DAE≅∠angle, D, A, E, \cong, angle

Reflexive property
10\dfrac{CA}{DA}=\dfrac{BA}{EA}
DA
CA
​
=
EA
BA
​
start fraction, C, A, divided by, D, A, end fraction, equals, start fraction, B, A, divided by, E, A, end fractionSubstitution (Part A, part B)
11\triangle CAB\sim \triangle DAE△CAB∼△DAEtriangle, C, A, B, \sim, triangle, D, A, E

similarity
Part D: Prove DE=\dfrac{1}{2}CBDE=
2
1
​
CBD, E, equals, start fraction, 1, divided by, 2, end fraction, C, B
[Got it, thanks!]
StatementReason
12\dfrac{CB}{DE}=\dfrac{CA}{DA}
DE
CB
​
=
DA
CA
​
start fraction, C, B, divided by, D, E, end fraction, equals, start fraction, C, A, divided by, D, A, end fractionLengths of corresponding pairs of sides of similar triangles have equal ratios. (Part C)
13\dfrac{CB}{DE}=2
DE
CB
​
=2start fraction, C, B, divided by, D, E, end fraction, equals, 2Substitution (Part A, 12)
14\dfrac{1}{2}CB=DE
2
1
​
CB=DEstart fraction, 1, divided by, 2, end fraction, C, B, equals, D, EMultiply both sides of the equation by \dfrac{DE}{2}
2
DE
​
start fraction, D, E, divided by, 2, end fraction. (13)