 Hello and welcome to the session, let us understand the following question which says, in a liquid lateral triangle prove that 3 times the square of one side is equal to 4 times the square of one of its altitudes. Here we have the following figure, triangle ABC which is an liquid lateral triangle and AD is perpendicular to BC. Now let us proceed on to the solution. Here in triangle ADB triangle ADC we have AB is equal to AC, sides of an equilateral triangle are equal to angle C because each angle of an equilateral triangle is 6 degrees. Angle ADB is equal to angle ADC equal to 90 degrees. Therefore, by RHS criterion of congruence we have triangle ADB congruent to triangle ADC. Also we know that corresponding parts of congruent triangles are equal. This implies VD is equal to DC. Also VD is equal to DC which is equal to half of BC. Let us name this as number 1. Since triangle ADB is a right triangle, right angle at D by Pythagoras theorem D square AB square is equal to AD square plus BD square or AB square is equal to AD square plus half BC the whole square by 1. This implies AB square is equal to AD square plus BC square by 4. This implies AB square is equal to AD square plus BC can be written as AB because all the sides of an equilateral triangle are equal. So we can write it as AB square by 4. This implies AB square minus AB square by 4 is equal to AD square. This implies 3 AB square by 4 is equal to AD square. This implies 3 AB square is equal to 4 AD square. That is we have proved 3 times the square of one side that is AB is the side of an equilateral triangle is equal to 4 times the square of one of its altitude and here AD is the altitude of the triangle ABC. So hence proved. I hope you have understood this question. That's all for this session. Bye and have a nice day.