Given: ab ii de, ac = ce
prove: a abc = a edc

Step-by-step explanation:

Here  given:   AB II DE, AC = CE

Now, in ΔABC and ΔEDC

As,  AB II DE (given)

⇒∠BAC = ∠CED   (Alternative pairs of angles)

and  ∠ABC = ∠CDE  (Alternative pairs of angles)

and AC = CE (given)

So, by the congruent  property of AAS

(ANGLE ANGLE SIDE) which states that two triangle are congruent if two angles and one side of one triangle is equal to the two angles and one side of the other triangle.

                                ΔABC ≅  ΔEDC

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°

in mathematics, a constraint is a condition of an optimization problem that the solution must satisfy. there are several types of constraints—primarily equality constraints, inequality constraints, and integer constraints. the set of candidate solutions that satisfy all constraints is called the feasible set.

hoped i'm right lol