Question

In the given figure, from a point O in the interior of triangle ABC, perpendiculars OD, OE, and OF are drawn to the sides BC, CA, and AB respectively. Prove that:
(a) $$A{F}^2+B{D}^2+C{E}^2=O{A}^2+O{B}^2+O{C}^2-O{D}^2-O{E}^2-O{F}^2$$
(b) $$A{F}^2+B{D}^2+C{E}^2=A{E}^2+C{D}^2+B{F}^2$$

Answer 1

Hint: Here, we need to prove the given equations. We will use linear pair angles property, and Pythagoras’s theorem in three right angles to get three equations. Then, we will use the three equations to prove the required equation. Similarly, we will prove the equation in part (b) using the linear pair property, Pythagoras’s theorem, and the equation proved in part (a).

Complete step-by-step answer:
(a)
We will use the Pythagoras’s theorem to prove that $$A{F}^2+B{D}^2+C{E}^2=O{A}^2+O{B}^2+O{C}^2-O{D}^2-O{E}^2-O{F}^2$$.
From the figure, we can observe that the triangles AOF, BOD, and AOE are right angled triangles.
In the right angled triangle AOF, AO is the hypotenuse, AF is the base, and OF is the perpendicular.
Using the Pythagoras’s theorem in right angled triangle AOF, we get
$$\Rightarrow A{O}^2=A{F}^2+O{F}^2$$
Subtracting $$O{F}^2$$ from both sides of the equation, we get
$$\Rightarrow A{F}^2=A{O}^2-O{F}^2$$
In the right angled triangle BOD, BO is the hypotenuse, BD is the base, and OD is the perpendicular.
Using the Pythagoras’s theorem in right angled triangle BOD, we get
$$\Rightarrow B{O}^2=B{D}^2+O{D}^2$$
Subtracting $$O{D}^2$$ from both sides of the equation, we get
$$\Rightarrow B{D}^2=B{O}^2-O{D}^2$$
From the figure, we can observe that the angles CEO and AEO form a linear pair.
Therefore, we get
$$\Rightarrow \mathrm{\angle }CEO+\mathrm{\angle }AEO={180}^{\circ }$$
Substituting $$\mathrm{\angle }AEO={90}^{\circ }$$ in the equation, we get
$$\Rightarrow \mathrm{\angle }CEO+{90}^{\circ }={180}^{\circ }$$
Subtracting $${90}^{\circ }$$ from both sides of the equation, we get
$$\Rightarrow \mathrm{\angle }CEO={90}^{\circ }$$
Therefore, the triangle COE is a right angled triangle.
In the right angled triangle COE, CO is the hypotenuse, CE is the base, and OE is the perpendicular.
Using the Pythagoras’s theorem in right angled triangle COE, we get
$$\Rightarrow C{O}^2=C{E}^2+O{E}^2$$
Subtracting $$O{E}^2$$ from both sides of the equation, we get
$$\Rightarrow C{E}^2=C{O}^2-O{E}^2$$
Now, adding both sides of the equations $$A{F}^2=A{O}^2-O{F}^2$$, $$B{D}^2=B{O}^2-O{D}^2$$, and $$C{E}^2=C{O}^2-O{E}^2$$, we get
$$A{F}^2+B{D}^2+C{E}^2=A{O}^2-O{F}^2+B{O}^2-O{D}^2+C{O}^2-O{E}^2$$
Rewriting the equation by rearranging the terms, we get
$$A{F}^2+B{D}^2+C{E}^2=O{A}^2+O{B}^2+O{C}^2-O{D}^2-O{E}^2-O{F}^2$$
Hence, we have proved that $$A{F}^2+B{D}^2+C{E}^2=O{A}^2+O{B}^2+O{C}^2-O{D}^2-O{E}^2-O{F}^2$$.
 (b)
From the figure, we can observe that the angles AFO and BFO form a linear pair.
Therefore, we get
$$\Rightarrow \mathrm{\angle }AFO+\mathrm{\angle }BFO={180}^{\circ }$$
Substituting $$\mathrm{\angle }AFO={90}^{\circ }$$ in the equation, we get
$$\Rightarrow \mathrm{\angle }BFO+{90}^{\circ }={180}^{\circ }$$
Subtracting $${90}^{\circ }$$ from both sides of the equation, we get
$$\Rightarrow \mathrm{\angle }BFO={90}^{\circ }$$
Therefore, the triangle BOF is a right angled triangle.
From the figure, we can observe that the angles BDO and CDO form a linear pair.
Therefore, we get
$$\Rightarrow \mathrm{\angle }BDO+\mathrm{\angle }CDO={180}^{\circ }$$
Substituting $$\mathrm{\angle }BDO={90}^{\circ }$$ in the equation, we get
$$\Rightarrow {90}^{\circ }+\mathrm{\angle }CDO={180}^{\circ }$$
Subtracting $${90}^{\circ }$$ from both sides of the equation, we get
$$\Rightarrow \mathrm{\angle }CDO={90}^{\circ }$$
Therefore, the triangle COD is a right angled triangle.
In the right angled triangle AOE, AO is the hypotenuse, AE is the base, and OE is the perpendicular.
Using the Pythagoras’s theorem in right angled triangle AOE, we get
$$\Rightarrow A{O}^2=A{E}^2+O{E}^2$$
Subtracting $$O{E}^2$$ from both sides of the equation, we get
$$\Rightarrow A{E}^2=A{O}^2-O{E}^2$$
In the right angled triangle BOF, BO is the hypotenuse, BF is the base, and OF is the perpendicular.
Using the Pythagoras’s theorem in right angled triangle BOF, we get
$$\Rightarrow B{O}^2=B{F}^2+O{F}^2$$
Subtracting $$O{F}^2$$ from both sides of the equation, we get
$$\Rightarrow B{F}^2=B{O}^2-O{F}^2$$
In the right angled triangle COD, CO is the hypotenuse, CD is the base, and OD is the perpendicular.
Using the Pythagoras’s theorem in right angled triangle COD, we get
$$\Rightarrow C{O}^2=C{D}^2+O{D}^2$$
Subtracting $$O{D}^2$$ from both sides of the equation, we get
$$\Rightarrow C{D}^2=C{O}^2-O{D}^2$$
We have proved that $$A{F}^2+B{D}^2+C{E}^2=O{A}^2+O{B}^2+O{C}^2-O{D}^2-O{E}^2-O{F}^2$$.
Grouping the terms using parentheses, we get
$$\Rightarrow A{F}^2+B{D}^2+C{E}^2=\left(O{A}^2-O{E}^2\right)+\left(O{B}^2-O{F}^2\right)+\left(O{C}^2-O{D}^2\right)$$
Rewriting the terms of the expression, we get
$$\Rightarrow A{F}^2+B{D}^2+C{E}^2=\left(A{O}^2-O{E}^2\right)+\left(B{O}^2-O{F}^2\right)+\left(C{O}^2-O{D}^2\right)$$
Substituting $$A{O}^2-O{E}^2=A{E}^2$$, $$B{O}^2-O{F}^2=B{F}^2$$, and $$C{O}^2-O{D}^2=C{D}^2$$ in the equation, we get
$$\Rightarrow A{F}^2+B{D}^2+C{E}^2=A{E}^2+B{F}^2+C{D}^2$$
Hence, we have proved that $$A{F}^2+B{D}^2+C{E}^2=A{E}^2+B{F}^2+C{D}^2$$.

Note: We used the sum of two angles in a linear pair in the solution. The sum of all the angles lying on a line is equal to $${180}^{\circ }$$. These angles are said to form a linear pair.
We used the Pythagoras’s theorem in the solution to solve the solution. The Pythagoras’s theorem states that the square of the hypotenuse of a right angled triangle is equal to the sum of squares of the other two sides, that is $$\text{Hypotenus}{\text{e}}^2=\text{Perpendicula}{\text{r}}^2+\text{Bas}{\text{e}}^2$$.