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

Given:

In a triangle $\triangle ABC, \angle A = 50^o, \angle B = 60^o$ and $\angle C = 70^o$.

To do:

We have to find the measures of the angles of the triangle formed by joining the mid-points of the sides of this triangle.

Solution:

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

$DE, EF$ and $ED$ are joined.

This implies,

$EF \parallel BC$

$DE \parallel AB$

$FD \parallel AC$

Therefore,

$BDEF$ and $CDEF$ are parallelograms.

This implies,

$\angle B = \angle E = 60^o$

$\angle C = \angle F = 70^o$

$\Rightarrow \angle A = \angle D = 50^o$

Hence, $\angle D = 50^o, \angle E = 60^o$ and $\angle F = 70^o$.

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Updated on: 10-Oct-2022

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