ABC and DBC are two triangles on the same base BC such that A and D lie on the opposite sides of BC, AB = AC and DB = DC. Show that AD is the perpendicular bisector of BC.
Solution:
Given, ABC and DBC are two triangles on the same base BC.
A and D lie on the opposite sides of BC
AB = AC
DB = DC
We have to show that AD is the perpendicular bisector of BC.
Considering triangle ABD and ACD,
Given, AB = AC
Common side = AD
Given, BD = CD
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 ABD and ACD are congruent.
By CPCT,
∠BAD = ∠CAD ------------ (1)
∠BAO = ∠CAO ------------ (2)
Considering triangles AOB and AOC,
Given, AB = AC
Common side = AO
From (2), ∠BAO = ∠CAO
By SAS criterion, the triangles AOB and AOC are congruent.
By CPCTC,
BO = OC
∠AOB = ∠AOC ------------ (3)
We know that the linear pair of angles is always equal to 180 degrees.
∠AOB + ∠AOC = 180°
From (3), ∠AOB + ∠AOB = 180°
2∠AOB = 180°
∠AOB = 180°/2
∠AOB = 90°
This implies AD is perpendicular to BC.
Therefore, AD is the perpendicular bisector of BC
✦ Try This: In given figure ABC and DBC are two isosceles triangles on the same base BC. Show that ∠ABD=∠ACD.
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 7
NCERT Exemplar Class 9 Maths Exercise 7.4 Problem 8
ABC and DBC are two triangles on the same base BC such that A and D lie on the opposite sides of BC, AB = AC and DB = DC. Show that AD is the perpendicular bisector of BC
Summary:
ABC and DBC are two triangles on the same base BC such that A and D lie on the opposite sides of BC, AB = AC and DB = DC. It is shown that AD is the perpendicular bisector of BC by CPCT
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