You visited us 2 times! Enjoying our articles? Unlock Full Access!

Any Point Equidistant from the End Points of a Segment Lies on the Perpendicular Bisector of the Segment

In t Δ ABC, ...

Question

In t $Δ$ ABC, AB > AC. E is the mid-point of BC and AD is perpendicular to BC. Then,
Answer: $A B^{2} + A C^{2} = 2 A E^{2} + 2 B E^{2}$
State whether the above statement is true=1 or false=0.

Open in App

Solution

Given: $△ A B C$ , $A D ⊥ B C$ , $B E = C E$
In $△ A B D$
$A D^{2} = A B^{2} - B D^{2}$ (Pythagoras theorem) (I)
In $△ A C D$
$A D^{2} = A C^{2} - C D^{2}$ (Pythagoras theorem) (II)
Adding I and II
$2 A D^{2} = A B^{2} + A C^{2} - B D^{2} - C D^{2}$
$2 A D^{2} = A B^{2} + A C^{2} - (B E + E D)^{2} - (C E - E D)^{2}$
$2 A D^{2} = A B^{2} + A C^{2} - B E^{2} - E D^{2} - 2 B E . E D - C E^{2} - E D^{2} + 2 C E . E D$
$2 A D^{2} = A B^{2} + A C^{2} - 2 B E^{2} - 2 E D^{2}$ (BE = CE)
$2 (A D^{2} + B E^{2} + D E^{2}) = A B^{2} + A C^{2}$
$2 (A E^{2} + B E^{2}) = A B^{2} + A C^{2}$ (Pythagoras Theorem in $△ A E D$ )

Suggest Corrections

Similar questions

A B C, A B > A C . E

B C

A D

B C

A B^{2} + A C^{2} = 2 A E^{2} + 2 B E^{2}

Δ A B C, A B = A C . D, E

F

B C, C A

A B

:

A D

F C .

Δ A B C

The perpendiculars from the mid-point of BC to AB and AC are equal.

If the above statement is true then mention answer as 1, else mention 0 if false.

Join BYJU'S Learning Program