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Question
In a triangle, ABC, AB = BC, AD is perpendicular to side BC and CE is perpendicular to side AB.
Prove that: AD = CE.
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Solution
ln ΔABD and ΔCBE,
AB = BC ....(given)
AD ⊥ BC
CE ⊥ AB
To proved:
In ΔABD & ΔCBE
∠ ADB = ∠ CEB = 90° ....[Perpendiculars]
∠B = ∠B ....(Common angle)
AB = BC
∴ ΔABD ≅ ΔCBE ....(by AAS congruence)
⇒ AD = CE ...(c.p.c.t.c)
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