1.AD bisects angle A of triangle ABC, where D lies on BC and angle C is greater than angle B. Prove that angle ADB is greater than angle ADC. 2.In a triangle ABC, right-angled at B, BD is drawn perpendicular to AC.Prove that: i) angle ABD = angle C ii)angle CBD = angle A

1
by mimansh

2015-04-08T13:59:28+05:30

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Angle DAB = angle DAC as AD is the bisector of angle A.

=  angle DAC + angle C
=  angle DAB + angle B
= angle DAC + angle B
if angle C is greater, then  angle ADB is greater.
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2.  angle ABD + angle DBC = 90
angle ABD = 90 - angle DBC
= 90 - (90 - angle C),  = angle C
as in right angle triangle DBC, angle BDC = 90, so angle DBC = 90 - angle C.
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similar to the above proof.

in right angle triangle ABD,  angle D = 90.
so angle ABD + angle A = 90
so angle ABD = 90 - angle A.

Now angle CBD =  90 - angle ABD = 90 - (90 - angle A)
= angle A.

i couldnt understand the 1st one
draw a diagram. and see. i hope you know what is meant by an exterior angle to a triangle ..
otherwise, just use : angle ADB = 180 - angle ADC. also, that the sum of three angles in a triangle is 180 deg,