In the given figure, AB=BC and angle ABO = angle CBO then prove that angle DAB = angle ECB
Answers
Answer:
∠DAB= ∠ECB (by CPCT)
Step-by-step explanation:
Given:-
AB=BC
∠ABO= ∠CBO
To Prove:-
∠DAB= ∠ ECB
Proof:-
In Δ ABO and ΔCBO;
AB=CB (given)
∠ABO= ∠CBO (i) (given)
OB=OB (common)
∴ Δ ABO ≅ Δ CBO (By SAS)
Now,
By CPCT;
∠AOB = ∠COB (ii)
∵Exterior angle is equal to the sum of two opposite units.
∴ ∠AOB + ∠ABO = ∠DAB
Also, ∠COB + ∠CBO= ∠ECB (iii)
As from eqⁿ (i) & (ii);
∠ABO= ∠CBO & ∠AOB = ∠COB
Also, LHS of both triangles are equal. So,
(∠ABO + ∠CBO) = (∠AOB + ∠COB)
=> ∠DAB = ∠ECB [ From eqⁿ (iii)]
Hence, ∠DAB= ∠ ECB
To know more about CPCT, you can visit the following links:-
https://brainly.in/question/1577784?referrer=searchResults
https://brainly.in/question/598068?referrer=searchResults
∠DAB = ∠ECB is proved by proving the congruence of ΔABO ≅ ΔCBO.
Given,
AB = BC
∠ABO = ∠CBO
To Prove,
∠DAB = ∠ECB
Solution,
We can solve this problem using a very simple method.
Let us consider ΔABO and ΔCBO.
We have been given that AB = BC.
We have also been given that ∠ABO = ∠CBO.
Also, OB is a common side for both triangles ΔABO and ΔCBO.
Hence, by SAS congruence criterion, ΔABO ≅ ΔCBO.
Now, since ΔABO ≅ ΔCBO, corresponding parts of congruent triangle (CPCT) are congruent.
Hence, ∠OAB = ∠OCB.
Let this be equation (1)
Now, since OE is a line segment, ∠OCB + ∠ECB = 180°
⇒ ∠ECB = 180° - ∠OCB
Let this be equation (2).
Similarly, since OD is a line segment, ∠OAB + ∠DAB = 180°
⇒ ∠DAB = 180° - ∠OAB
Let this be equation (3).
From equation (1), ∠OAB = ∠OCB.
Multiply by -1 on both sides: - ∠OAB = - ∠OCB
Add 180° on both sides: 180° - ∠OAB = 180° - ∠OCB
From equations (2) and (3): ∠DAB = ∠ECB.
Hence proved that ∠DAB = ∠ECB.
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