Given: AD = BC and AD || BC Prove: ABCD is a parallelogram.
2 answers:
You might be interested in
Answer:
(3/4)a
Step-by-step explanation:
The angle at K is 120°, so the angle at L is its supplement: 60°. That makes triangle FKL an equilateral triangle with a base of FL = a. The vertex at K is centered over the base, so is a/2 from G.
The midsegement length is the average of GK and FL, so is ...
midsegment = (GK +FL)/2 = (a/2 +a)/2
midsegment = (3/4)a
Answer:
In isosceles triangle ABC, BM is the median to the base AC and Point D is on BM as shown below in the figure;
Median of a triangle states that a line segment joining a vertex to the midpoint of the opposing side, bisecting it
M is the median of AC
then by definition;
AM = MC ......[1]
In ΔAMD and ΔDMC
AM = MC [side] [By [1]]
[Angle]
DM =DM [Common side]
Side-Angle-Side postulate(SAS) states that if two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then these two triangles are congruent.
Then, by SAS,
CPCT stands for Corresponding parts of congruent triangles are congruent
By CPCT,
[Corresponding side] ......[2]
In ΔABD and ΔCBD
AB = BC [Side] [By definition of isosceles triangle]
BD= BD [common side]
AD = DC [Side] [by [2]]
Side-Side-Side(SSS) postulate states that if three sides of one triangle are congruent to three sides of another triangle, then the two triangles are congruent.
Therefore, by SSS theorem,
