O is a point in the interior of a square ABCD such that OAB is an equilateral triangle. Show that ∆ OCD is an isosceles triangle.
Solution:
Given, O is a point in the interior of a square ABCD such that OAB is an equilateral triangle.
We have to show that OCD is an isosceles triangle.
Join OC and OD.
Given, AOB is an equilateral triangle
∠OAB = ∠OBA = 60° ------------------ (1)
We know that each angle of a square is equal to 90 degrees.
∠DAB = ∠CBA = 90° ------------------------ (2)
Subtracting (1) and (2),
∠DAB - ∠OAB = ∠CBA - ∠OBA
From the figure,
∠DAB - ∠OAB = ∠DAO
∠CBA - ∠OBA = ∠CBO
So, ∠DAO = ∠CBO = 90° - 60° = 30°
Considering triangles AOD and BOC,
Sides of equilateral triangle, AO = BO
∠DAO = ∠CBO
Sides of a square, AD = BC
SAS criterion states that if two sides of one triangle are proportional to two sides of another triangle and their included angles are congruent, then the triangles are congruent.
By SAS criterion, the triangles AOD and BOC are congruent
Considering triangle COD,
By CPCTC,
OC = OD
Therefore, OCD is an isosceles triangle.
✦ Try This: Two isosceles triangle are shown below. If 180° - z = 2y and y = 75°, what is the value of x?
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 7
NCERT Exemplar Class 9 Maths Exercise 7.4 Problem 7
O is a point in the interior of a square ABCD such that OAB is an equilateral triangle. Show that ∆ OCD is an isosceles triangle
Summary:
O is a point in the interior of a square ABCD such that OAB is an equilateral triangle. It is shown that ∆ OCD is an isosceles triangle by CPCT
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