In the given figure: $O P, O Q$ and $O R$ are drawn perpendicular to the sides $A B C, C A$ and $A B$ respectively of triangle $A B C$ . Prove that
${A F}^{2} + {B P}^{2} + {C Q}^{2} = {A Q}^{2} + {C P}^{2} + {B R}^{2}$

Open in App

Solution

In $△ O A R, O A^{2} = O R^{2} + A R^{2}$
In $△ O A Q, O A^{2} = O Q^{2} + A Q^{2}$
$⟹ O R^{2} + A R^{2} = O Q^{2} + A Q^{2} . . . . . . (1)$
Similarly $O P^{2} + B P^{2} = O R^{2} + B R^{2} . . . . . . . (2), C Q^{2} + O Q^{2} = C P^{2} + O P^{2} . . . . . . (3)$
$(1) + (2) + (3) ⟹ A R^{2} + B P^{2} + C Q^{2} = A Q^{2} + C P^{2} + B R^{2}$

Suggest Corrections

Similar questions

Q. In the given figure,

O P, O Q

and

O R

are drawn perpendiculars to the sides

B C, C A

and

A B

respectively of triangle

A B C

. Prove that:

{A R}^{2} + {B P}^{2} + {C Q}^{2} = {A Q}^{2} + {C P}^{2} + {B R}^{2}

Q. In given figure from a point O in the interior of a

△ A B C

, perpendiculars OD, OE and OF are drawn to the sides BC, CA and AB respectively. Prove that :

(i)

{A F}^{2} + {B D}^{2} + {C E}^{2} = {O A}^{2} + {O B}^{2} + {O C}^{2} - {O D}^{2} - {O E}^{2} - {O F}^{2}

(ii)

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

Q. In a right triangle ABC right-angled at B, if P and Q are points on the sides AB and AC respectively, then

(a) AQ² + CP² = 2(AC²+ PQ²)
(b) 2(AQ² + CP²) = AC² + PQ²
(c) AQ²+ CP² = AC² + PQ²
(d)

AQ + CP = \frac{1}{2} (AC + PQ)

P and Q are points on sides CA and CB respectively of Δ ABC right angled at C. Prove that ${A Q}^{2}$ + ${B P}^{2}$ = ${A B}^{2}$ + ${P Q}^{2}$ [4 MARKS]

State true or false

In a triangle ABC, P and Q are the id-points of AC and AB respectively and angle

B A C = 90^{o}

.
then

B P^{2} + C Q^{2} = 5 P Q^{2}

Join BYJU'S Learning Program

In the given figure: OP,OQ and OR are drawn perpendicular to the sides ABC,CA and AB respectively of triangle ABC. Prove that AF2+BP2+CQ2=AQ2+CP2+BR2

In △OAR,OA2=OR2+AR2In △OAQ,OA2=OQ2+AQ2⟹OR2+AR2=OQ2+AQ2......(1)Similarly OP2+BP2=OR2+BR2.......(2),CQ2+OQ2=CP2+OP2......(3)(1)+(2)+(3)⟹AR2+BP2+CQ2=AQ2+CP2+BR2

In the given figure: $O P, O Q$ and $O R$ are drawn perpendicular to the sides $A B C, C A$ and $A B$ respectively of triangle $A B C$ . Prove that
${A F}^{2} + {B P}^{2} + {C Q}^{2} = {A Q}^{2} + {C P}^{2} + {B R}^{2}$