In figure given below, D and E are mid-points of the sides $B C$
and CA respectively of a
$△ A B C$ , right angled at $C$ .
Prove that:
(i) $4 {A D}^{2} = 4 {A C}^{2} + {B C}^{2}$
(ii) $4 {B E}^{2} = 4 {B C}^{2} + {A C}^{2}$
(iii) $4 ({A D}^{2} + {B E}^{2}) = 5 {A B}^{2}$

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Solution

Proof:
ACD is a right triangle.
So, ${A D}^{2} = {A C}^{2} + {C D}^{2}$ [Pythagoras theorem]
Multiply both sides by 4 , get:
$4 {A D}^{2} = 4 {A C}^{2} + 4 {C D}^{2}$
$4 {A D}^{2} = 4 {A C}^{2} + 4 {B D}^{2} [$ As: $D$ is the midpoint of $B C, C D = B D$ $= \frac{1}{2} B C]$
$4 {A D}^{2} = 4 {A C}^{2} + (2 B D)^{2}$
$4 {A D}^{2} = 4 {A C}^{2} + {B C}^{2}$
... (1) $[A s : B C = 2 B D]$
(ii) BCE is a right triangle.
${B E}^{2} = {B C}^{2} + {C E}^{2}$
[Pythagoras theorem]
Multiply both sides by 4 , get:
$4 {B E}^{2} = 4 {B C}^{2} + 4 {C E}^{2}$
$4 {B E}^{2} = 4 {B C}^{2} + (2 C E)^{2}$
$4 {B E}^{2} = 4 {B C}^{2} + {A C}^{2}$
$\dots$ (2) [As: E is the mid-point of
$A C, A E = C E = \frac{1}{2} A C]$
(iii) Adding equation (1) and (2)
$4 {A D}^{2} + 4 {B E}^{2} = 4 {A C}^{2} + {B C}^{2} + 4 {B C}^{2} + {A C}^{2}$
$4 {A D}^{2} + 4 {B E}^{2} = 5 {A C}^{2} + 5 {B C}^{2}$
$4 ({A D}^{2} + {B E}^{2}) = 5 ({A C}^{2} + {B C}^{2})$
$4 ({A D}^{2} + {B E}^{2}) = 5 ({A B}^{2})$

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D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE²+ BD² = AB² + DE²

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