Question

8.
seg
In figure 3.63, seg AD I
BC.
seg AE is the bisector of ZCAB and
E
Pe
C-E-D.
Prove that
ZDAE
1
2
- (ZC- ZB)
B
A
Fig. 3.63​

Answer 1

Answer:

Given.

AE is bisector of angle BAC, hence

angle BAE = angle EAC = X LET.........(A)

Now, AD is perpendicular to BC

hence angle ADC = angle ADB = 90°

now in ∆ ADC

(I am not using the word angle now)

DAC + DCA + ADC = 180°

DAC + DCA = 180 - ADC = 189-90 = 90°

DAC + DCA = 90°...................(1)

similarly from ∆ ADB

ABD + BAD = 90°...................(2)

(1) - (2)

DAC + DCA - ABD - BAD = 0

DCA - ABD = BAD - DAC

or

BAD - DAC = C - B. (DCA = C, ABD = B)

BAE + EAD - (EAC - EAD) = C - B

X + EAD - X + EAD = C - B. (from A)

2*EAD = C - B

EAD = (1/2)*(C - B)

Answer 2

Answer:

Step-by-step explanation:

Given.

AE is bisector of angle BAC, hence

angle BAE = angle EAC = X LET.........(A)

Now, AD is perpendicular to BC

hence angle ADC = angle ADB = 90°

now in ∆ ADC

(I am not using the word angle now)

DAC + DCA + ADC = 180°

DAC + DCA = 180 - ADC = 189-90 = 90°

DAC + DCA = 90°...................(1)

similarly from ∆ ADB

ABD + BAD = 90°...................(2)

(1) - (2)

DAC + DCA - ABD - BAD = 0

DCA - ABD = BAD - DAC

or

BAD - DAC = C - B. (DCA = C, ABD = B)

BAE + EAD - (EAC - EAD) = C - B

X + EAD - X + EAD = C - B. (from A)

2*EAD = C - B

EAD = (1/2)*(C - B)