Question

1. In ∆ABC the angle bisectors drawn from vertexes A and B intersect at point D. Find ∠ADB if: m∠A = α , m∠B = β

2.Angle bisectors of the angles ∠A and ∠B of a triangle ∆ABC intersect each other at point M. Find m∠AMB, if m∠A = 58°, m∠B = 96°.

Answer 1

Answer (1)

As given in question

In ∆ABC,  angle bisectors drawn from vertexes A and B intersect at point D

m∠A = α , m∠B = β

Find ∠ADB

To proof

As shown in the diagram given below

AE and BF is angle bisector.

( angle bisector are the bisector which divided angle in two equal parts )

m∠A = α , m∠B = β   ( given)

Thus

$\angle BAD = \frac{\alpha }{2} \\ \angle ABD =\frac{\beta}{2}$

Now in ΔADB

∠ADB +∠ BAD +∠ ABD = 180 °

( Angle sum property of a triangle )

$\angle ADB = 180^{\circ}- \frac{\alpha }{2}-\frac{\beta }{2}$

Hence proved

ANSWER (2)

.Angle bisectors of the angles ∠A and ∠B of a triangle ∆ABC intersect each other at point M.

m∠A = 58°, m∠B = 96°

Find m∠AMB

To proof

As shown in the diagram given below

AE and BF is angle bisector

$\angle BAM = \frac{\ 58 }{2} \\ \angle ABM =\frac{\ 96}{2}$

∠BAM = 29°

∠ABM = 48°

InΔAMB

∠AMB +∠BAM +∠ABM = 180°

( Angle sum property of a triangle)

putting the value

we get

∠AMB = 180 - 48 - 29

∠AMB = 103 °

Hence proved