AAS Postulate
It is given that CE = BD so we know "S" (representing side) has to be in the three letter postulate.
It is also given that angle DBA and angle CEA are right angles, so therefore they are congruent. Now we know that an "A" must also be in the postulate.
Lastly, we know that the triangles have a second angle, EAB, in common because they share it overlappingly. So there must be another "A" in the postulate.
Now we need to look at the order in which it is presented. The order follows Angle, Angle, Side so the postulate must be the AAS postulate. Hope this helps!
It will be a parallelogram.
Answer:
m∠JKP = 31.5°
Step-by-step explanation:
Incenter of a triangle is the point where all the bisectors of interior angles intersect each other.
JN is the angle bisector of ∠KJL.
Therefore, m∠KJN = m∠LJN
(7x - 6) = (5x + 4)
7x - 5x = 6 + 4
2x = 10
x = 5
m∠KJN = (7x - 6)
= 7(5) - 6
= 35 - 6
= 29°
In ΔKJN,
m∠JKN + m∠KNJ + m∠NJK = 180°
m∠JKN + 90° + 29° = 180°
m∠JKN = 180°- 119° = 61°
Since KO is the angle bisector of ∠JKN,
m∠JKP = 
= 
= 30.5°
Answer: 8:05
Step-by-step explanation:
1/3 * 7/9 kg = (1 * 7)/(3 * 9) kg = 7/27 kg
He used 7/27 kg of sugar.