Question

ABC is an isosceles triangle in which altitude BE and CF are drawn to equal sides AC and AB, respectively. Show that \[\vartriangle ABE \cong \vartriangle ACF\]

Answer 1

Hint: An altitude is always perpendicular from the base to the opposite vertex.
A triangle is a polygon with three edges and three vertices, which are the basic shapes in geometry. It is a closed two-dimensional shape with three straight sides.
A triangle has three sides, and their type depends on the length of its sides and the size of its angles. There are basically three types of a triangle based on the length of the sides, namely: Scalene Triangle, Isosceles Triangle, and Equilateral Triangle.
An isosceles triangle is a triangle that has two sides of equal length. All the three angles of the isosceles triangle are acute angle, i.e., less than,\[{90^ \circ }\]whereas it’s the total sum of internal angle is \[{180^ \circ }\]

Complete step-by-step answer:
Given \[\vartriangle ABC\]is an isosceles triangle
Also \[AC = AB - - - (i)\]
Line BE is drawn from vertex B, and it joins the side AC, and a line CF is drawn from vertex C, which joins the side AB,
So BE and CF are the altitude, and as we know, a line drawn from a line to the opposite vertex is always perpendicular at its base.
Therefore,
\[\angle AEB = \angle AFC = {90^ \circ } - - - (ii)\]
Now in triangle \[\vartriangle AEB\]and \[\vartriangle AFC\]
\[\angle AEB = \angle AFC\] [By equation (ii)]
\[\angle A = \angle A\][Common angle]
BE=CF [Isosceles triangle]
Since three of the condition AAS for \[\vartriangle AEB\]and \[\vartriangle AFC\]satisfies hence we can conclude \[\vartriangle ABE \cong \vartriangle ACF\]
Hence proved

Note: Students must note that whenever an angle for two triangles is common, then the angle is known as a common angle, and it fulfills one condition to equate two triangles.