Question

In triangle ABC,AD bisects angle BAC and AD= DC.if the angle BDA =70° , then the measurement of angle ABD is ?

Answer 1

given: AD=DC
Hence angle dac=angle dca =x(say)
we know that the exterior angle is equal to sum of its opposite interior angle
hence, angle ADB=angle (dac+dca)
70=x+ x
x=35
and angle bad=angle dac = 35°
Now, in triangle ABD we have
70°+35°+angle ABD=180°
angle ABD=180°-105°=75°

Answer 2

Answer:

75°

Step-by-step explanation:

Refer the attached figure

Since AD bisects ∠BAC

So, 2∠BAD = 2∠DAC = ∠BAC

AD= DC

Opposite angle of equal sides are equal

So, $\angle DAC = \angle DCA$ ---1

So, In ΔADC

$\angle DAC+ \angle DCA =\angle ADB$ (Exterior angle property)

$2\angle DAC =70$

$\angle DAC =35$

So,  $\angle DAC = \angle DCA =35$

∠DAC = 2∠BAC

2(35) = ∠BAC

70= ∠BAC

In  ΔABC

∠A+∠B+∠C=180 (Angle sum property )

70+∠B+35=180

∠B=180 -105

∠B=75

Hence The measurement of angle ABD is 75°