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Let ABCD be a trapezium in which AB || DC nd let E be the mid point of AD. let F be a point on BC such that EF || AB. Prove that (i) F is midpoint of BC, (ii) EF = 1/2(AB DC).?
Verified Answer
Let ABCD be a trapezium in which AB || DC nd let E be the mid point of...
Figure shows the trapezium ABCD.
AB || CD (given)
E is mid point of AD (given)
EF || DC (given)
Let us join BD and let EF meets BD at G.
since EF || DC and AB || DC, we have EG || AB.
In Δ ABD, EG || AB and E is mid point of AD. Hence EG = (1/2)AB and BG = DG ..(1)
similarly we can say GF || DC.
in Δ BCD, FG || CD and G is mid point of BD. Hence GF = (1/2)DC ..(2)
from (1) and (2), we have EG+GF = EF = (1/2)(AB+DC)
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Most Upvoted Answer
Let ABCD be a trapezium in which AB || DC nd let E be the mid point of...
Proof:
Let's break down the proof into two parts:
(i) F is the midpoint of BC
(ii) EF = 1/2(AB + DC)
(i) F is the midpoint of BC:
To prove that F is the midpoint of BC, we need to show that BF = FC.
Since AB || DC, we can conclude that triangle ABE is similar to triangle DCE by the AAA similarity criterion.
As E is the midpoint of AD, we can also conclude that triangle ABE is similar to triangle CDE by the SAS similarity criterion.
From the above two similarities, we can deduce that triangle ABE is similar to triangle DCE.
Since EF || AB, we can use the corresponding angles formed by parallel lines to establish that triangle BEF is also similar to triangle CDE.
By the similarity of triangle BEF and triangle CDE, we can conclude that angles BFE and CED are equal.
Now, let's consider triangle BCF and triangle CDE.
Since angles BFE and CED are equal, and angles BFC and CDE are corresponding angles formed by parallel lines BF and DE, we can conclude that triangle BCF is also similar to triangle CDE.
By the similarity of triangle BCF and triangle CDE, we can establish that angles BCF and CDE are equal.
Since angles BCF and CDE are equal, we can conclude that angles BCF and BFC are also equal.
Therefore, triangle BCF is an isosceles triangle, and BF = FC.
Hence, F is the midpoint of BC.
(ii) EF = 1/2(AB + DC):
To prove that EF = 1/2(AB + DC), we will use the fact that F is the midpoint of BC.
Since F is the midpoint of BC, we can conclude that BF = FC.
Let's denote AB as a and DC as b.
By the similarity of triangle BEF and triangle CDE, we can establish the following ratios:
EF/DE = BE/CE
Since E is the midpoint of AD, we can substitute DE with a/2 and CE with b/2:
EF/(a/2) = BE/(b/2)
Simplifying the above equation, we get:
EF = (a/2) * (BE/CE)
Since AB || DC, we can conclude that BE/CE = AB/DC = a/b.
Substituting a/b for BE/CE in the equation EF = (a/2) * (BE/CE), we get:
EF = (a/2) * (a/b) = a^2 / (2b)
Similarly, we can establish that CF = (b/2) * (a/b) = a^2 / (2b)
Since BF = FC, we can conclude that BF = CF = a^2 / (2b).
Therefore, EF = 1/2(AB + DC).
Hence, we have proved that EF = 1/2(AB + DC).
In conclusion, we have proved both (i) F is the midpoint of BC and (ii) EF = 1/2(AB + DC) by using the properties of similar triangles and the fact that F is the midpoint of BC.
Let's break down the proof into two parts:
(i) F is the midpoint of BC
(ii) EF = 1/2(AB + DC)
(i) F is the midpoint of BC:
To prove that F is the midpoint of BC, we need to show that BF = FC.
Since AB || DC, we can conclude that triangle ABE is similar to triangle DCE by the AAA similarity criterion.
As E is the midpoint of AD, we can also conclude that triangle ABE is similar to triangle CDE by the SAS similarity criterion.
From the above two similarities, we can deduce that triangle ABE is similar to triangle DCE.
Since EF || AB, we can use the corresponding angles formed by parallel lines to establish that triangle BEF is also similar to triangle CDE.
By the similarity of triangle BEF and triangle CDE, we can conclude that angles BFE and CED are equal.
Now, let's consider triangle BCF and triangle CDE.
Since angles BFE and CED are equal, and angles BFC and CDE are corresponding angles formed by parallel lines BF and DE, we can conclude that triangle BCF is also similar to triangle CDE.
By the similarity of triangle BCF and triangle CDE, we can establish that angles BCF and CDE are equal.
Since angles BCF and CDE are equal, we can conclude that angles BCF and BFC are also equal.
Therefore, triangle BCF is an isosceles triangle, and BF = FC.
Hence, F is the midpoint of BC.
(ii) EF = 1/2(AB + DC):
To prove that EF = 1/2(AB + DC), we will use the fact that F is the midpoint of BC.
Since F is the midpoint of BC, we can conclude that BF = FC.
Let's denote AB as a and DC as b.
By the similarity of triangle BEF and triangle CDE, we can establish the following ratios:
EF/DE = BE/CE
Since E is the midpoint of AD, we can substitute DE with a/2 and CE with b/2:
EF/(a/2) = BE/(b/2)
Simplifying the above equation, we get:
EF = (a/2) * (BE/CE)
Since AB || DC, we can conclude that BE/CE = AB/DC = a/b.
Substituting a/b for BE/CE in the equation EF = (a/2) * (BE/CE), we get:
EF = (a/2) * (a/b) = a^2 / (2b)
Similarly, we can establish that CF = (b/2) * (a/b) = a^2 / (2b)
Since BF = FC, we can conclude that BF = CF = a^2 / (2b).
Therefore, EF = 1/2(AB + DC).
Hence, we have proved that EF = 1/2(AB + DC).
In conclusion, we have proved both (i) F is the midpoint of BC and (ii) EF = 1/2(AB + DC) by using the properties of similar triangles and the fact that F is the midpoint of BC.
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Let ABCD be a trapezium in which AB || DC nd let E be the mid point of AD. let F be a point on BC such that EF || AB. Prove that (i) F is midpoint of BC, (ii) EF = 1/2(AB DC).?
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Let ABCD be a trapezium in which AB || DC nd let E be the mid point of AD. let F be a point on BC such that EF || AB. Prove that (i) F is midpoint of BC, (ii) EF = 1/2(AB DC).? for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about Let ABCD be a trapezium in which AB || DC nd let E be the mid point of AD. let F be a point on BC such that EF || AB. Prove that (i) F is midpoint of BC, (ii) EF = 1/2(AB DC).? covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let ABCD be a trapezium in which AB || DC nd let E be the mid point of AD. let F be a point on BC such that EF || AB. Prove that (i) F is midpoint of BC, (ii) EF = 1/2(AB DC).?.
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