In triangle abc,D is the mid point of ab and e is the mid point of ac such that de parallel bc then, AE:AC IS???
Answers
Given:
In ΔABC, D is a midpoint of AB
DE // BC
To find:
The ratio of AE : ED
Solution:
We know that,
Converse of Basic Proportionality theorem:- If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
∴ ....... (i)
Since D is a midpoint of AB (given)
∴ AD = DB ...... (ii)
From (i) & (ii), we get
AE = EC
⇒ E is a midpoint of AC ..... (iii)
Now,
Consider Δ AED and ΔACB, we have
∠A = ∠A ........ [common angle]
∠ADE = ∠ABC ..... [corresponding angles, ∵DE//BC and side AB forms a transversal]
∴ Δ AED ~ ΔACB ........ By AA Similarity
Also, we know that the corresponding sides of two similar triangles are proportional to each other.
∴ AE/AC=ED/BC
on rearranging, we get
⇒ AE/ED=AC/BC
⇒ AE/ED = (AE+EC)/BC
⇒ AE : ED = (AE+EC) : BC
HOPE YOU ARE REALLY HELPED
Given :-
In ΔABC, D is a midpoint of AB
DE // BC
To find :-
The ratio of AE : ED
Solution :-
We know that,
Converse of Basic Proportionality theorem:- If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
∴ AD / DB = AE / EC ....... (i)
Since D is a midpoint of AB (given)
∴ AD = DB ...... (ii)
From (i) & (ii), we get
AE = EC
⇒ E is a midpoint of AC ..... (iii)
Now,
Consider Δ AED and ΔACB, we have
∠A = ∠A ........ [common angle]
∠ADE = ∠ABC ..... [corresponding angles, ∵DE//BC and side AB forms a transversal]
∴ Δ AED ~ ΔACB ........ By AA Similarity
Also, we know that the corresponding sides of two similar triangles are proportional to each other.
∴ AE / AC = ED / BC
on rearranging, we get
⇒ AE / ED = AC / BC
⇒ AE / ED = AE + EC / BC
⇒ AE : ED = (AE+EC) : BC
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