harish and deep were trying to prove a theorem for this they did the following.. 1) Drew a triangle ABC 2) D and E are found the mid point of AB and AC
3) DE was joined and DE is extended to F so DE =EF
4) FC is joined Answer the following question angle EFC is congruent to which angle?
Answers
Given:
D and E are found the midpoint of AB and AC
DE =EF
To find:
i. ∆ADE and ∆EFC congruency criteria
ii. Angle EFC is equal to which angle?
iii. Angle ECF is equal to which angle?
iv. CF is equal to which line?
v. CF is parallel to which line?
Solution:
∆ADE and ∆EFC are congruent by SAS rule, angle EFC= angle ADE, angle ECF= angle DAE, CF=BD, and CF is parallel to BD.
We can find the solution by following the given process-
i. In ∆ADE and ∆EFC,
DE=EF (Given)
Angle AED= Angle CEF (Vertically opposite angles)
AE=EC (E is the midpoint of AC)
So, the ∆ADE and ∆EFC are congruent by the SAS congruency rule. (Option c)
ii. Again, in ∆ADE and ∆EFC, we know that
AD and CF are two parallel lines and DF is a transversal.
So, angle EFC= angle ADE (Alternate interior angles) (Option b)
iii. Since AD and CF are parallel and AC is also a transversal between the two lines,
angle ECF= angle DAE (Alternate interior angles) (Option a)
iv. ∆ADE and ∆EFC are congruent triangles.
So, their corresponding sides and angles will also be equal.
CF=AD (∆ADE and ∆EFC are congruent)
Since D is the midpoint of AB, AD=DB.
Thus, CF=BD. (Option a)
v. Since DE=EF, CF is parallel to AB.
So, CF is parallel to AB, AD, BD.
CF is parallel to BD. (Option iii)
Therefore, ∆ADE and ∆EFC are congruent by SAS rule, angle EFC= angle ADE, angle ECF= angle DAE, CF=BD, and line CF is parallel to BD.
Answer:
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