What else would need to be congruent to show that AABC=A DEF by the AAS theorem?
8
Given:
ܪ ܂
ZA ZD
2B RPE
F
A. AC = DF
B. 2Cz2p
C. BE ZE
D. BC = DF

D. AC ≅ DF

Step-by-step explanation:

According to the AAS Theorem, two triangles are considered congruent to each other when two angles and a mon-included side of one triangle are congruent to two corresponding angles and a corresponding non-included side of the other.

Thus, in the diagram given:

<A and <B in ∆ABC are congruent to corresponding angles <D and <E in ∆DEF.

The only condition left to be met before we can conclude that both triangles are congruent by the AAS Theorem is for a mon-included side AC to be congruent to corresponding non-included side DF.

So, AC ≅ DF is what is needed to make both triangles congruent.

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°