A square is inscribed in an isosceles right triangle so that the square and the triangle have one angle common. Show that the vertex of the square opposite the vertex of the common angle bisects the hypotenuse.
Solution:
Given, a square is inscribed in an isosceles right triangle.
The square and the triangle have one angle in common.
We have to show that the vertex of the square opposite the vertex of the common angle bisects the hypotenuse.
Consider an isosceles right triangle ABC right angled at A.
A square DEF is inscribed in the triangle.
Given, ∠A = 90°
We know that in an isosceles triangle two sides have equal length.
So, AB=AC ------------------------ (1)
We know that all sides of a square are equal
So, AD = AF ---------------------- (2)
On subtracting (1) and (2), we get
AB - AD = AC - AF
BD = CF ------------------------ (3)
Considering triangles CFE and BDE,
The sides of a square DE = EF
From (3), BD = CF
∠CFE = ∠EDB = 90°
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 similar.
By SAS criteria, the triangles CFE and BDE are similar.
The Corresponding Parts of Congruent Triangles are Congruent (CPCTC) theorem states that when two triangles are similar, then their corresponding sides and angles are also congruent or equal in measurements.
By CPCTC,
CE = BE
Therefore, vertex E of the square bisects the hypotenuse BC.
✦ Try This: In a right angled triangle, if one angle is 45° and the side opposite to it is K times the hypotenuse, then the value of K is
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 8
NCERT Exemplar Class 9 Maths Exercise 8.4 Problem 1
A square is inscribed in an isosceles right triangle so that the square and the triangle have one angle common. Show that the vertex of the square opposite the vertex of the common angle bisects the hypotenuse
Summary:
A square is inscribed in an isosceles right triangle so that the square and the triangle have one angle in common. It is shown that the vertex of the square opposite the vertex of the common angle bisects the hypotenuse
☛ Related Questions:
- In a parallelogram ABCD, AB = 10 cm and AD = 6 cm. The bisector of ∠A meets DC in E. AE and BC produ . . . .
- P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD . . . .
- P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD . . . .
visual curriculum