the line segment joining the midpoints of two sides of a triangle is parallel to the third side and half of it
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> Given: A △ABC in which D and E are the mid-points of side AB and AC respectively. DE is joined .
>To Prove: DE || BC and DE = 1/2 BC.
> Construction: Produce the line segment DE to F , such that DE = EF. Join FC
> Proof : In △s AED and CEF,
AE = CE [∵ E is the mid point of AC]
∠AED = ∠CEF [Vertically Opposite angles]
DE = EF [by Construction]
∴ △AED ≅ △CEF [by SAS congruence axiom]
⇒ AD = CF ----(1) [c.p.c.t.]
& ∠ADE and ∠CEF ----(2) [c.p.c.t.]
Now, D is the mid point of AB.
⇒ AD = DB ----(3)
From (1) and (3), CF = DB ----(4)
Also, from (2)
⇒ AD || FC [if a pair of alternate interior angless are equal then lines are parallel]
⇒ DB || BC ----(5)
From (4) and (5), we find that DBCF is a quadrilateral such that one pair of opposite sides are equal and parallel.
∴ DBCF is a Parallelogram
⇒ DF || BC and DF = BC [∵ Opp. side of parallelogram are equal and parallel]
Also, DE = EF [by construction]
Hence, DE || BC and DE = 1 / 2 BC
Same Question's Answer: > Given: A △ABC in which D and E are the mid-points of side AB and AC respectively. DE is joined .
>To Prove: DE || BC and DE = 1/2 BC.
> Construction: Produce the line segment DE to F , such that DE = EF. Join FC
> Proof : In △s AED and CEF,
AE = CE [∵ E is the mid point of AC]
∠AED = ∠CEF [Vertically Opposite angles]
DE = EF [by Construction]
∴ △AED ≅ △CEF [by SAS congruence axiom]
⇒ AD = CF ----(1) [c.p.c.t.]
& ∠ADE and ∠CEF ----(2) [c.p.c.t.]
Now, D is the mid point of AB.
⇒ AD = DB ----(3)
From (1) and (3), CF = DB ----(4)
Also, from (2)
⇒ AD || FC [if a pair of alternate interior angless are equal then lines are parallel]
⇒ DB || BC ----(5)
From (4) and (5), we find that DBCF is a quadrilateral such that one pair of opposite sides are equal and parallel.
∴ DBCF is a Parallelogram
⇒ DF || BC and DF = BC [∵ Opp. side of parallelogram are equal and parallel]
Also, DE = EF [by construction]
Hence, DE || BC and DE = 1 / 2 BC
Same Question here: https://brainly.in/question/2433243
Cheers!
> Given: A △ABC in which D and E are the mid-points of side AB and AC respectively. DE is joined .
>To Prove: DE || BC and DE = 1/2 BC.
> Construction: Produce the line segment DE to F , such that DE = EF. Join FC
> Proof : In △s AED and CEF,
AE = CE [∵ E is the mid point of AC]
∠AED = ∠CEF [Vertically Opposite angles]
DE = EF [by Construction]
∴ △AED ≅ △CEF [by SAS congruence axiom]
⇒ AD = CF ----(1) [c.p.c.t.]
& ∠ADE and ∠CEF ----(2) [c.p.c.t.]
Now, D is the mid point of AB.
⇒ AD = DB ----(3)
From (1) and (3), CF = DB ----(4)
Also, from (2)
⇒ AD || FC [if a pair of alternate interior angless are equal then lines are parallel]
⇒ DB || BC ----(5)
From (4) and (5), we find that DBCF is a quadrilateral such that one pair of opposite sides are equal and parallel.
∴ DBCF is a Parallelogram
⇒ DF || BC and DF = BC [∵ Opp. side of parallelogram are equal and parallel]
Also, DE = EF [by construction]
Hence, DE || BC and DE = 1 / 2 BC
Same Question's Answer: > Given: A △ABC in which D and E are the mid-points of side AB and AC respectively. DE is joined .
>To Prove: DE || BC and DE = 1/2 BC.
> Construction: Produce the line segment DE to F , such that DE = EF. Join FC
> Proof : In △s AED and CEF,
AE = CE [∵ E is the mid point of AC]
∠AED = ∠CEF [Vertically Opposite angles]
DE = EF [by Construction]
∴ △AED ≅ △CEF [by SAS congruence axiom]
⇒ AD = CF ----(1) [c.p.c.t.]
& ∠ADE and ∠CEF ----(2) [c.p.c.t.]
Now, D is the mid point of AB.
⇒ AD = DB ----(3)
From (1) and (3), CF = DB ----(4)
Also, from (2)
⇒ AD || FC [if a pair of alternate interior angless are equal then lines are parallel]
⇒ DB || BC ----(5)
From (4) and (5), we find that DBCF is a quadrilateral such that one pair of opposite sides are equal and parallel.
∴ DBCF is a Parallelogram
⇒ DF || BC and DF = BC [∵ Opp. side of parallelogram are equal and parallel]
Also, DE = EF [by construction]
Hence, DE || BC and DE = 1 / 2 BC
Same Question here: https://brainly.in/question/2433243
Cheers!
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The line-segment joining the mid-points of any two sides of a triangle is parallel to the third side and is half of it. ... The quadrilateral formed by joining the mid-points of the sides of a quadrilateral, in order, is a parallelogram
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I think so
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