Who is a point in the interior of a square ABCD such that AB is an equilateral triangle so that triangle Cod is an isosceles triangle?

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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We don’t have your requested question, but here is a suggested video that might help.

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.

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.