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Area of a Triangle

In Δ ABC,AB=...

Question

In $Δ$ ABC,AB=AC and AD is a median. Prove that $A D ⊥ B C$ .

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Solution

We have,
$I n Δ A B C, A B = A C$ .and $A D$ is the median.
Then prove that,
$I n Δ A B C, A D$ is the perpendicular bisector of $B C$ .
Proof:-
In
$Δ A D B a n d Δ A D C$
$A B = A C (G i v e n t h a t)$
$A D = A D (C o m m o n l i n e)$
$∠ A D B = ∠ A D C (E v e r y r i g h t a n g l e)$
Then, $Δ A D B ≅ Δ A D C$ by S.A.S. rule
So, $A D$ is the perpendicular bisector of $B C$ .
Hence proved.

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