ABCD is quadrilateral such that AB = AD and CB = CD. Prove that AC is the perpendicular bisector of BD.
Solution:
Given, ABCD is a quadrilateral
AB = AD
CB = CD
We have to prove that AC is the perpendicular bisector of BD.
Considering triangles ABC and ADC,
Common side = AC
Given, AB = AD
BC = BD
SSS Criterion (Side-Side-Side) states that if all the three sides of one triangle are equal to the three corresponding sides of another triangle, the two triangles are said to be congruent.
By SSS criterion, the triangles ABC and ADC are congruent
The Corresponding Parts of Congruent Triangles are Congruent (CPCTC) theorem states that when two triangles are congruent, then their corresponding sides and angles are also congruent or equal in measurements.
By CPCTC,
∠1 = ∠2
Considering triangles AOB and AOD,
Common side = AO
Given, AB = AD
∠1 = ∠2
By SSS criterion, the triangles AOB and AOD are congruent
By CPCTC,
BO = DO
∠3 = ∠4
We know that the linear pair of angles is always supplementary.
So, ∠3 + ∠4 = 180°
∠3 + ∠3 = 180°
2∠3 = 180°
∠3 = 180°/2
∠3 = 90°
This implies that AC is perpendicular to BD
Therefore, AC is the perpendicular bisector of BD.
✦ Try This: In given figure, If AD⊥BC and BD/DA = DA/DC, prove that △ABC is a right triangle.
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 7
NCERT Exemplar Class 9 Maths Exercise 7.4 Problem 21
ABCD is quadrilateral such that AB = AD and CB = CD. Prove that AC is the perpendicular bisector of BD
Summary:
ABCD is quadrilateral such that AB = AD and CB = CD. It is proven that AC is the perpendicular bisector of BD. The perpendicular bisectors of a triangle are lines passing through the midpoint of each side which are perpendicular to the given side
☛ Related Questions:
- O is a point in the interior of a square ABCD such that OAB is an equilateral triangle. Show that ∆ . . . .
- ABC and DBC are two triangles on the same base BC such that A and D lie on the opposite sides of BC, . . . .
- ABC is an isosceles triangle in which AC = BC. AD and BE are respectively two altitudes to sides BC . . . .