Name: Figures on the Same Base
Uploaded: 2019-07-19
Duration: 6 min 34 s
Description: Two figures are said to be on one base and between the same parallels if they have a common side as the base and certain vertices lie on the line parallel to...

Question

Perpendiculars are drawn from a point within an equilateral triangle to the three sides. Prove that the sum of the three perpendiculars is equal to the altitude of the triangle.

Accepted Answer

Let $△ A B C$ is an equilateral triangle. $P$ is any point in the interior of the triangle.

Let $h$ be the altitude of the equilateral triangle $A B C$ .

$P D ⊥ B C, P E ⊥ A C$ and $P F ⊥ A B$ are drawn.

We have to prove $h = P D + P E + P F$ .

We know that the area of triangle is given by $\frac{1}{2} \times base \times height$ .

So, we have, $ar (△ A B C) = \frac{1}{2} \times h \times B C$ -----(i)

Similarly, we have, $ar (△ A P B) = \frac{1}{2} \times P F \times A B = \frac{1}{2} \times P F \times B C$ -----(ii) ( $AB = BC$ ).

Also, $ar (△ A P C) = \frac{1}{2} \times P E \times A C = \frac{1}{2} \times P E \times B C$ -----(iii) ( $AC = BC$ )

And, $ar (△ B P C) = \frac{1}{2} \times P D \times B C$ -----(iv)

Adding equations (ii), (iii) and (iv) we get, $ar (△ A P B) + ar (△ A P C) + ar (△ B P C) = \frac{1}{2} \times P F \times B C + \frac{1}{2} \times P E \times B C + \frac{1}{2} \times P D \times B C$

$\Rightarrow ar (△ A B C) = \frac{1}{2} \times B C \times (P F + P E + P D)$

$\Rightarrow \frac{1}{2} \times B C \times h = \frac{1}{2} \times B C \times (P F + P E + P D)$ [From equation (i)]

$\Rightarrow h = P F + P E + P D$

Hence, proved.

Perpendiculars are drawn from a point within an equilateral triangle to the three sides. Prove that the sum of the three perpendiculars is equal to the altitude of the triangle.

Important Questions on Theorems on Area

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