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Question
ABC and DBC are two isosceles triangles on the same base BC (see the given figure). Show that ∠ABD = ∠ACD.
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Solution
In isosceles △ABC,
AB = AC ...[ABC is an isosceles triangle]
∴ ∠ACB = ∠ABC …(i) ...[Angles opposite to equal sides of a triangle are equal]
Also, in isosceles △BCD,
BD = DC ...[BDC is an isosceles triangle]
∴ ∠BCD = ∠CBD ...(ii) ...[Angles opposite to equal sides of a triangle are equal]
On adding the corresponding sides of (i) and (ii)
∠ACB + ∠BCD = ∠ABC + ∠CBD
⇒ ∠ACD = ∠ABD or ∠ABD = ∠ACD ...(By Corresponding parts of congruent triangles)
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