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Question
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.
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Solution
Given in the question, 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.
To proof that AD is the perpendicular bisector of BC that is OB = OC.
Proof: In triangle BAD and triangle CAD,
AB = AC ...[Given]
BD = CD ...[Given]
AD = AD ...[Common side]
Now, by SSS criterion of congruence,
ΔBAD ≅ ΔCAD
So, ∠1 = ∠2 ...[CPCT]
Now, in triangle BAO and triangle CAO,
AB = AC ...[Given]
∠1 = ∠2 ...[Proved above]
AO = AO ...[Common side]
So, by SAS criterion of congruence,
ΔBAO ≅ ΔCAO
Since, BO = CO ...[CPCT]
And ∠3 = ∠4 ...[CPCT]
∠3 + ∠4 = 180° ...[Linear pair axiom]
∠3 + ∠3 = 180°
2∠3 = 180°
∠3 = `(180^circ)/2`
∠3 = 90°
Therefore, AD is perpendicular to bisector of BC.
Hence proved.
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