Question

o is any point outside the equilateral triangle abc and within the angular region on abc ; op,oq,or are the perpendicular distance of ab,bc, and ca respectively from the point o. prove that the altitude of the triangle=op+oq-or​

Answer 1

Answer:

Step-by-step explanation:

Area of triangle ABC = Area of triangle OCB + triangle OAC – triangle OAB

if O is point out side of AB

Area of triangle ABC = (1/2) * Any side * Altitude

Area of triangle OCB = (1/2) BC * OQ

Area of triangle OAC = (1/2) AC * OR

Area of triangle OAB = (1/2) AB * OP

AB = BC = AC  ( equilateral Triangle)

=>   (1/2) * Any side * Altitude = (1/2) BC * OQ +  (1/2) AC * OR -   (1/2) AB * OP

=> (1/2) * AB * Altitude = (1/2)* AB ( OQ + OR - OP)

=> Altitude = OQ + OR - OP

This means point should be outside  AC  to get desired results

then

Area of triangle ABC = Area of triangle OCB + triangle OAB – triangle OAC

if O is point out side of AC

Area of triangle ABC = (1/2) * Any side * Altitude

Area of triangle OCB = (1/2) BC * OQ

Area of triangle OAC = (1/2) AC * OR

Area of triangle OAB = (1/2) AB * OP

AB = BC = AC  ( equilateral Triangle)

=>   (1/2) * Any side * Altitude = (1/2) BC * OQ +  (1/2) AC * OP -   (1/2) AB * OR

=> (1/2) * AB * Altitude = (1/2)* AB ( OQ + OP - OR)

=> Altitude = OQ + OP - OR

=>  Altitude = OP + OQ - OR