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Question
In an isosceles triangle ABC, sides AB and AC are equal. If point D lies in base BC and point E lies on BC produced (BC being produced through vertex C), prove that:
(i) AC > AD
(ii) AE > AC
(iii) AE > AD
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Solution
We know that the bisector of the angle at the vertex of an isosceles triangle bisects the base at the right angle.
Using Pythagoras theorem in AFB,
AB2 = AF2 + BF2 ...(i)
In AFD,
AD2 = AF2 + DF2 ...(ii)
We know ABC is isosceles triangle and AB = AC
AC2 = AF2 + BF2 ..(iii)[ From (i)]
Subtracting (ii) from (iii)
AC2 - AD2 = AF2 + BF2 - AF2 - DF2
AC2 - AD2 = BF2 - DF2
Let 2DF = BF
AC2 - AD2 = (2DF)2 - DF2
AC2 - AD2 = 4DF2 - DF2
AC2 = AD2 + 3DF2
⇒ AC2 > AD2
⇒ AC > AD
Similarly, AE > AC and AE > AD.
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