Prove: The segment joining the midpoints of two sides of a triangle is parallel to the third side.
1 answer:
Answer:
As per theorem Segment joining midpoints of two sides of a triangle is parallel to the third side.
Step-by-step explanation:
Given : In a triangle ABC, D and E are the midpoint of sides AB and AC.
To Proof : DE║BC
Construction: A line CF parallel to BA is drawn and De is extended to point F.
Proof : In a triangle ADE and EFC
Sides AD║CF and DF is transverse.Then ∠ADE = ∠CFE (Alternate angles)
Similarly AD║CF and AC is transverse then ∠DAE = ∠FCA (Alternate angles)
Therefore ΔADE≅ΔCEF (AAS property)
From this property side AD = Side CF (corresponding sides of the congruent triangles)
Therefore BD = CF
So sides BD and CF are equal and parallel to each other which represents DBCF is a parallelogram.
Therefore DE║BC
Hence proved.
You might be interested in
Answer:
-13 -
+ 4p + 3
Step-by-step explanation:
I am having a little trouble reading the number. I think 5 is the question number and not part of the problem. I think this is the problem:
(4p - 6 - 3) - (8
+ 7
- 6) You take the opposite of everything inside the second set of parentheses. You can think of it as multiplying through by -1
4p - 6 - 3 - 8
- 7
+ 6 Now we combine like terms. That is the numbers with the same variable p and the same power of p.
-13 - 8
+ 4p + 3
Answer:
angle y is 64⁰
Step-by-step explanation:
There are 2 ways to solve this.
first,
they are alternate angles
second,
you can solve by
180-64=116 (on a straight line)
180-116=64⁰
I'm not really good at explaining but hope this helps :)
Unfortunately, my answer could not be saved. Please refer to the attachments below.
