Question

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.

Answer 1

$\huge \boxed{\bf{\red{Solution:-}}}$

$\bf \green{Given:-} \: \sf \: D \: and \: E \: are \: points \: on \: the \\ \: \: \: \: \: \: \: \: \: \: \: \: \sf sides \: CA \: and \: CB \: respectively \: of \: a \: triangle \: ABC \\ \: \: \: \: \: \: \: \: \: \: \: \: \sf right \: angled \: at \: C.$

$\bf \green{To \: Prove:-} \: \sf \: AE {}^{2} + BD {}^{2} = AB {}^{2} + DE {}^{2} .$

$\bf \green{Proof:-} \: \sf \: In \: right \: triangle \: ACB,$

$\: \: \: \: \: \: \: \: \: \: \: \: \sf \: AB {}^{2} = \: AC {}^{2} + BC {}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ...(1)$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf\big| \: \: \: \: By \: Pythagoras \: Theorem$

$\sf \: In \: right \: triangle \: DCE,$

$\: \: \: \: \: \: \sf \: DE {}^{2} = \: CD {}^{2} + CE {}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:... (2)$

$\sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \big| \: \: By \: Pythagoras \: Theorem$

$\sf \: Adding \: (1) \: and \: (2) ,\: We \: get$

$\sf \: RHS = AB {}^{2} + DE {}^{2}$

$\: \: \: \: \: \: \: \: \sf = (AC {}^{2} + BC {}^{2} ) + (CD {}^{2} + CE {}^{2} )$

$\: \: \: \: \: \: \: \sf = (AC {}^{2} + CE {}^{2} ) + (BC {}^{2} + CD {}^{2} )$

$\: \: \: \: \: \ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf...(3)$

$\sf \: Now \: in \: right \: triangle \: ACE,$

$\sf \: AC {}^{2} + CE {}^{2} = AE {}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ...(4)$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf(By \: Pythagoras \: Theorem)$

$\sf \: and \: in \: right \: triangle \: BCD,$

$\sf \: BC {}^{2} + CD {}^{2} = BD {}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:... (5)$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf(By \: Pythagoras \: Theorem)$

$\sf \: Putting \: values \: from \: (4) \: and \: (5) \: in \: (3), \:$

$\sf \: w e\: have,$

$\sf \: RHS = AB {}^{2} + DE {}^{2} = AE {}^{2} + BD {}^{2} = LHS.$

Answer 2

$\large\underline\red{GIVEN:-}$

$\textbf{D and E are the points on the sides}$$\textbf{CA and CB respectively of a triangle}$$\textbf{ABC and right angled at C}$

$\large\underline\red{TO \: PROVE:-}$

${AE^2+BD^2=AB^2+DE^2}$

$\large\underline\blue{NOTE:-}$

$\bold\green{(Pythagoras \: theorem)}$

$\textbf{In a right triangle the square of}$$\textbf{hypotenuse is equal to}$$\textbf{the square of other two sides}$

${(hypotenuse)^2=(base)^2+(perpendicular)^2}$

$\large\underline\red{PROOF:-}$

$\textbf{According to Pythagoras theorem}$

$\bold\green{In \: triangle \: DCE}$

${DE^2=DC^2+CE^2}$ ------(i)

$\bold\green{In \: triangle \: ABC}$

${AB^2=AC^2+CB^2}$ ------(ii)

$\bold\green{In \: triangle \: ACE}$

${AE^2=AC^2+CE^2}$ ------(iii)

$\bold\green{In \: triangle \: BCD}$

${BD^2=DC^2+CB^2}$-------(iv)

$\textbf{Adding (iii) and (iv)}$

${AE^2+BD^2=(AC^2+CE^2)+(DC^2+CB^2)}$

${AE^2+BD^2=AC^2+CE^2+DC^2+CB^2}$

${AE^2+BD^2=(AC^2+CB^2)+(CE^2+DC^2)}$

[using (i) and (ii) ]

${AE^2+BD^2=AB^2+DE^2}$

Hence,

AE²+BD² = AB²+DE²----- proved