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.
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The most appropriate choice for similarity of triangles will be given by -
Speed of tip of the shadow of woman = 6 ft/s
What are similar triangles?
Two triangles are said to be similar, if the corrosponding angles of the triangles are same and the corrosponding sides of the triangles are in the same ratio.
Here,
The diagram has been attached here
Let the distance of woman from the pole be x ft and the distance of tip of the shadow to the pole be y ft.
Height of street light = 18 ft
Height of woman = 6ft
The two triangles are similar [As height of woman is parallel to the height of pole]
To find the speed, we have to differentiate both sides with respect to time 't'
Speed of tip of her shadow = 6 ft
To learn more about similarity of triangles, refer to the link-
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