Question

In a triangle $\angle ABC,\angle A={50}^{\circ },\angle B={60}^{\circ }and\angle C={70}^{\circ }$. Find the measures of the angles of the triangle formed by joining the mid -points of the sides of this triangle.

Answer 1

.In $\Delta ABC,$

$\angle A={50}^{\circ },\angle B={60}^{\circ }and\angle C={70}^{\circ }$

D, E and F are the mid points of sides BC, CA and AB respectively

DE, EF and FD are joined

$\because$ D, E and F are the mid points of sides BC, CA and AB respectively

$\therefore$ EF || BC

DE || AB and FD || AC

$\therefore$ BDEF and CDEF are parallelogram.

$\therefore \angle B=\angle E={60}^{\circ }and\angle C=\angle F={70}^{\circ }$

Similar way, DE || AF, so AEDF forms a parallelogram.
So, $\angle A=\angle D={50}^{\circ }$

Hence $\angle D={50}^{\circ },\angle E={60}^{\circ }and\angle F={70}^{\circ }$