Draw a triangle $A B C$ in which $∠ A = 120^{o}$ and $A B = A C = 3 c m$ . Draw the bisectors angle $A$ .

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Solution

$! (B A D) =! (D A C) = 60^{0}$
$= 60^{0}$

$1.$ Draw a line segment $A B = 3$ cm
$2.$ Using protractor from $B$ measure from $0^{\circ}$ and mark a point at $120^{\circ}$
$3.$ From the point $A$ using compass draw an arc of length $3$ cm
$∴ A B = A C = 3$ cm
$4.$ Join $B C$
$5.$ Draw an arc at $∠ A$
$6.$ Bisect the arc at $∠ A$ and name the point of intersection of the arc as $D$
$7.$ Join $A D$
$t h e r e f o r e A D$ is the angle bisector.
Hence $∠ B A D = ∠ D A C = 60^{\circ}$
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Draw a triangle ABC in which ∠A=120o and AB=AC=3 cm. Draw the bisectors angle A.

Draw a triangle $A B C$ in which $∠ A = 120^{o}$ and $A B = A C = 3 c m$ . Draw the bisectors angle $A$ .