Question

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

Answer 1

Hint:
$\Delta ABC$ is the isosceles triangle. Use the property that angles opposite to equal sides are equal. After that, use the SAS congruence rule. Try it, you will get the answer.

Complete step by step solution:
Here it is given that $\Delta ABC$ is an isosceles triangle.
It is given that, $AB=AC$.
Now we know the property that angles opposite to equal sides are equal.
Now $AB$ and $AC$ are opposite and equal sides of $\angle ABD$ and $\angle ACE$. So, we get,
$\angle ABD=\angle ACE$ ……… (1)
Also, we are given that, $BE=CD$
Now subtracting $DE$ from both sides we get,
$BE-DE=CD-DE$
From the figure we can say that, $BE-DE=BD$ and $CD-DE=CE$.
So, we get,
$BD=CD$ …………………. (2)
Now taking $\Delta ABD$ and $\Delta ACE$,
$AB=AC$ ………… (given)
$\angle ABD=\angle ACE$ ……… (From (1))
$BD=CD$ …………………. (From (2))
Therefore, by SAS congruence rule we get,
 $\Delta ABD \cong \Delta ACE$
Now we get, $\Delta ABD \cong \Delta ACE$.
$\therefore AD=AE$
Hence proved.

Additional information:
As the two sides are equal in this triangle, the unequal side is called the base of the triangle. The angles opposite to the two equal sides of the triangle are always equal. The altitude of an isosceles triangle is measured from the base to the vertex (topmost ) of the triangle. A right isosceles triangle has the third angle as 90 degrees.


Note:
An isosceles triangle is a triangle which has any two of its sides equal to each other. Also, the angles opposite these equal sides are equal. In general, a triangle is a polygon which has three sides and three vertices. The sides and angles of the triangle could vary. SAS congruence rule is ‘Two triangles are said to be congruent if two sides and the included angle of one triangle are equal to the two sides and the included angle of the other triangle.