Question

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Answer 1

2answer).

Angle ADC=angel A+ angle B+angle C

join AC, we have two triangle

∆ABC and ∆ACD

In ∆ABC,

angle ABC+angle BAC+angle BCA———(eq.1)

In ∆ADC,

angle ADC+angle DCA+angle CAD———(eq.2)

angle BCA=angle BCD+angle DCA

angle BAC=angle BAD+angle DAC

from eq(1) and (2)

angle ABC+angle BAC+angle BCA=angel ADC+angel BCA+angle CAD

angle ABC+angle BAD+angle DAC=angle ADC+angle DCA+angle CAD

(angle DCA+angle DAC=angle DCA+angle CAD this angle are cancel)

angle ABC+angle BAC+angle BCD=angle ADC

angle B+angle A+angle C=angle ADC

therefore, angel ADC=angle A+angle B+angle C

5answer).

QT is the bisector of angle PQR and RI is the bisector of angle PQR.

angle TQR=1\2angle PQR and angle TRS=angle 1\2angle PRS———(eq.1)

angle PRS is an exterior angle of ∆PQR

angle PRS=angle QPR+angle PQR

1\2angle PRS=1\2angle QPR+1\2angle PQR

angle TRS=1\2angle QPR+angle TQR(by(1))———(eq.2)

In∆TQR, angle TRS is an exterior angle.

therefore, angle TRS=angle QTR+TQR———(eq.3)

from(2) and (3), we get

angle QTR+angle TQR=1\2angle QPR+angle TQR

therefore, angle QTR=1\2angle QPR

3answer).

angles forms a point is equal to 360°

(x+y)=(w+z)

x+y+x+y=360°

2x+2y=360°

2(x+y)=360°

x+y=360°\2

x+y=180°

w+z=180°

therefore, angel X angle Y forms a linear pair

therefore, AOB is a line

4answer).

ray BO is the bisector of angle CBE

angle CBO=angle EBO=1\2angle CBE

angle CBO=1\2(180°-y)

angle CBO=90°-y\2

ray CO is bisector of angle of BCD

angle BCO=angle OCD=1\2angle BCD

angle BCO=1\2(180°-z)

angle BCO=90°-z\2

In ∆BCO,

angle BCO+angle BOC+angle CBO=180°

90°-z\2+angle BCO+90°-y\2=180°

angle BOC=z\2+y\2

angle BOC=y+z\2———(eq.1)

In ∆ABC,

angle A+angle B+angle C=180°

x+y+z=180°

y+z=180°-x———(eq.2)

angle BCO=18-x\2

angle BCO=90°-x\2

angle BCO=90°-angle BAC\2

hope it is helpful....

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