Question

in the figure, D is the mid-point of side BC of a AABC and ZABD = 50°. if
AD=BD = CD, then find the measure of ZACD.​

Answer 1

Answer:

angle BAD=50 BDA is a issosles triangle

so angle ADC=50+50=100(by exterior angle property)

angle DAC=DCA(because it is a issosles triangle)

so 80÷2=40

Step-by-step explanation:

thanks

Answer 2

Answer:if angle B= 50.given = AD=BD=CD.

angle B=angle A. (property of isoscelus triangle)

50=50.in triangle ABD.

Angle ABD+BAD+ADB=180 degree. (angle sum property of triangle) 50+50+ADB=180.

100+ADB=180.

ADB=80. degree. ------> i

In triangle ABC.

angle ADB + angle ADC =180.

80 +ADC=180.ADC=100.--------> ii

In triangle ADC.

angle A=angle C.

and angle ADC+DCA+CAD =180.

100 +2(DCA)=180.DCA=40 degree.

so angle ACD =40 DEGREE.

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