Question

squares AMNB and AOPC are drawn on the sidesof triangle ABC so that they lie outside the triangle, prove that MC=OB

Answer 1

Given: squares AMNB and AOPC are drawn on the sides of triangle ABC.

To find:  prove that MC=OB.

Solution:

• As we have given that squares lie outside the triangle, so lets join MC and OB.

• Now,  we can see that:

           ang MAB = ang OAC   (as both the angles are 90°)

• So from this, we can conclude that:

          ang MAB + ang BAC = ang OAC + ang BAC

• By this we can conclude that:

           ang MAC = ang OAB .....................(i)

• Now lets consider triangle MAC and triangle BAO, we get:

           BA=MA (they are sides of a square) .................(ii)

           AO=AM (they are also sides of a square) ............(iii)

• So, from Eq(1), Eq(2) and Eq(3)

        triangle MAC ≅ triangle BAO (SAS or Side Angle Side)

Answer:

           Then, MC=OB (corresponding sides of two congruent triangles)