The median to the base of an isosceles triangle is:

Perpendicular to base

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Inclined

60^{\circ}

to the base

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Inclined

45^{\circ}

to the base

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Inclined

30^{\circ}

to the base

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A Perpendicular to base
Let $A B C$ be an isosceles triangle in which $A B = A C$
Let $\to A$ be the origin of reference and let $- - \to A B = \to b, - - \to A C = \to c$
Let $D$ be the middle point of $B C$
Then $- - \to A D = \frac{\to b + \to c}{2}$ and $- - \to B C = \to c - \to b ∴ - - \to A D . - - \to B C = \frac{\to b + \to c}{2} . (\to c - \to b) = \frac{1}{2} (| \to c |^{2} - | \to b |^{2}) = \frac{1}{2} (| A C |^{2} - | A B |^{2}) = 0$
Since the two sides are of equal length. Hence, median to the base of an isosceles triangle is perpendicular to the base of the triangle.

Suggest Corrections