Question
In a triangle $\angle\text{ABC},$ $\angle\text{A}=50^\circ,$ $\angle\text{B}=60^\circ$ andFind the measures of the angles of the triangle formed by joining the mid-points of the sides of this triangle.
✓
Answer
In $\triangle\text{ABC},$
$D$ and $E$ are mid points of $AB$ and $BC$.
By Mid point theorem, $\text{DE}||\text{AC},\ \text{DE}=\frac{1}{2}\text{AC}$ $F$ is the midpoint of $AC$ Then,
$\text{DE}=\Big(\frac{1}{2}\Big)\text{AC}=\text{CF}$ In a Quadrilateral $DECF$
$\text{DE}||\text{AC},\ \text{DE}=\text{CF}$
Hence $DECF$ is a parallelogram.
$\therefore\angle\text{C}=\angle\text{D}=70^\circ$ [Opposite sides of a parallelogram]
Similarly $BEFD$ is a parallelogram, $\angle\text{B}=\angle\text{F}=60^\circ$ $ADEF$ is a parallelogram,
$\angle\text{A}=\angle\text{E}=50^\circ$
$\therefore$ Angles of $\triangle\text{DEP}$ are: $\angle\text{D}=70^\circ,\angle\text{E}=50^\circ,\angle\text{F}=60^\circ$
$D$ and $E$ are mid points of $AB$ and $BC$.
By Mid point theorem, $\text{DE}||\text{AC},\ \text{DE}=\frac{1}{2}\text{AC}$ $F$ is the midpoint of $AC$ Then,
$\text{DE}=\Big(\frac{1}{2}\Big)\text{AC}=\text{CF}$ In a Quadrilateral $DECF$
$\text{DE}||\text{AC},\ \text{DE}=\text{CF}$
Hence $DECF$ is a parallelogram.
$\therefore\angle\text{C}=\angle\text{D}=70^\circ$ [Opposite sides of a parallelogram]
Similarly $BEFD$ is a parallelogram, $\angle\text{B}=\angle\text{F}=60^\circ$ $ADEF$ is a parallelogram,
$\angle\text{A}=\angle\text{E}=50^\circ$
$\therefore$ Angles of $\triangle\text{DEP}$ are: $\angle\text{D}=70^\circ,\angle\text{E}=50^\circ,\angle\text{F}=60^\circ$
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