Question 7 Triangles - Exercise 7.4
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Answer:
According to the question,
It is given that: A square ABCD and OA = OB = AB.
To prove:
Δ OCD is an isosceles triangle.
Proof:
In square ABCD,
Since ∠1 and ∠2 is equal to 90o
∠1 = ∠2 …(1)
Now, in Δ OAB , we have
Since ∠3 and ∠4 is equal to 60o
∠3 = ∠4 …(2)
Subtracting equations (2) from (1),
We get
∠1−∠3 = ∠2 −∠4
⇒ ∠5 = ∠6
Now,
In Δ DAO and Δ CBO,
AD = BC [Given]
∠5 = ∠6 [Proved above]
OA = OB [Given]
By SAS criterion of congruence,
We have
Δ DAO ≅ Δ CBO
OD = OC
⇒ Δ OCD is an isosceles triangle.
Hence, proved.
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In square $A B C D$ (not shown), point $E$ lies in the interior of $A B C D$ in such a way that $\triangle A B E$ is an equilateral triangle. Find $\mathrm{m} \angle D E C$
Given: O is a point in the interior of a square ABCD such that ΔOAB is an equilateral triangle.
Construction: Join OC and OD.
To show: ΔOCD is an isosceles triangle.
Proof: Since, AOB is an equilateral triangle.
∴ ∠OAB = ∠OBA = 60° ......(i)
Also, ∠DAB = ∠CBA = 90° ....(ii) [Each angle of a square is 90°] [∵ ABCD is a square]
On subtracting equation (i) from equation (ii), we get
∠DAB – ∠OAB = ∠CBA – ∠OBA = 90° – 60°
i.e. ∠DAO = ∠CBo = 30°
In ΔAOD and ΔBOC,
AO = BO .......[Given] [All the side of an equilateral triangle are equal]
∠ADO = ∠CBO ......[Proved above]
And AD = BC .....[Sides of a square are equal]
∴ ΔAOD ≅ ΔBOC ......[By SAS congruence rule]
Hence OD = OC ......[By CPCT]
In ΔCOD, OC = OD
Hence, ΔCOS is an isosceles triangle.
HEnce proved.