Question

In triangle ABC, Prove that if altitudes on three sides are equal, then triangle is equilateral.​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that AD, BE and CF be the altitudes drawn from A, B and C on BC, AC and AB respectively intersecting BC, AC and AB at D, E, F respectively.

Now, it is given that, altitudes on three sides are equal.

So, AD = BE = CF = h (say)

Now,

$\rm \: Area _{(\triangle\:ABC)} = \dfrac{1}{2} \times AD \times BC = \dfrac{1}{2} \times h \times BC - - - (1)$

Also,

$\rm \: Area _{(\triangle\:ABC)} = \dfrac{1}{2} \times BE \times AC =\dfrac{1}{2} \times h \times AC - - - (2)$

Also,

$\rm \: Area _{(\triangle\:ABC)} = \dfrac{1}{2} \times CF \times AB = \dfrac{1}{2} \times h \times AB - - - (3)$

Now, from equation (1), (2) and (3), we get

$\rm \: \dfrac{1}{2} \times h \times \: AB = \dfrac{1}{2} \times h \times BC = \dfrac{1}{2} \times h \times AC$

$\rm\implies \:AB = BC = AC$

$\rm\implies \: \triangle\:ABC \: is \: an \: equilateral \: triangle.$

$\rule{190pt}{2pt}$

Additional Information :-

$\color{green}\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}$

Answer 2

Step-by-step explanation:

Given : In △ ABC , AD, BE and CF are the altitudes drawn from A, B and C on BC, CA and AB respectively. and A D = B E = C F

To prove : △ ABC is an equilateral triangle.

Proof : In △ BEC and △ BFC , we have

Hyp. BE = CF (given)

Side BC = BC (common)

$\\ \therefore \: \: \Delta BEC \cong \triangle BFC \: \: \: \: \: \: \: \: \: \: \text{(RHS axiom of congrency)}$

$\text{\( \therefore \angle C =\angle B \) \quad (c.p.c.t.)}$

$\text{\( \therefore A B = A C \quad \) (sides opposite to equal angles)...(i)}$

Similarly, in right △ AFC and △ ADC, we have

$\begin{array}{rlr} & \text { Hyp. } A C=A C & \text { (common) } \\ & C F=A D & \text { (given) } \\ \therefore & \triangle A F C \cong \triangle A D C \quad \text { (R.H.S. axiom of congruency) } \\ \therefore & \angle A=\angle C \\ \therefore & B C=A B \quad \text { (sides opposite to equal angles) ...(ii) } \\ & \text { from (i) and (ii), we have } \\ & A B=B C=C A\end{array}$

Hence,△ABC is an equilateral triangle.