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# In an isosceles triangle ABC with AB = AC, D and E are points on BC such that BE = CD (see Fig.). Show that AD = AE.

Last updated date: 13th Jun 2024
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Hint:
1) In a triangle, the angles opposite to the equal sides are also equal and vice versa.
2) Two triangles are congruent to each other in any of the following ways:
SAS: If two of the sides and the angle between them are equal to the corresponding sides and angle of another triangle, then the triangles are congruent.
ASA: If two angles and the side between them are equal to the corresponding angles and side of another triangle, then the triangles are congruent.
SSS: When all the three sides are the same then the triangles are congruent.

Complete step by step solution:
Since the triangle is isosceles with AB = AC, the angles opposite to these sides must also be equal.
⇒ ∠ B = ∠ C.

Now, in Δ ABE and Δ ADC:
AB = AC (given)
BE = DC (given)
∠ B = ∠ C (by property)
∴ By the SAS congruence theorem, Δ ABE ≅ Δ ADC.
So, AE = AE because they are opposite to the equal angles ∠ B and C.
Hence, proved.

Note:
1) There is an SSA congruence as well, which states that two triangles are congruent if two of their corresponding sides and an angle NOT between them are also equal, provided, the angle MUST BE 90˚ or more than 90˚. If the angle is acute, there may be either a single case (congruent) or two possible cases (congruent in one and not necessarily congruent in the other one).
2) The sides (a, b, c) and the angles opposite to them (A, B, C) of a triangle ABC are connected by the sine formula, i.e. $\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}$ , which enables one to conclude that greater the angle, greater is the side opposite to it, and equal angles have equal sides opposite to them.