Question

in the given figure triangle ABC.if AB=AC and BD=DC,then find the measure of angle ADC

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Answer 1

Answer:90

90 ∘

90 ∘

90 ∘ Given that, in △ABC,

90 ∘ Given that, in △ABC,AB=AC, BD=DC

90 ∘ Given that, in △ABC,AB=AC, BD=DC To find out: ∠ADC

90 ∘ Given that, in △ABC,AB=AC, BD=DC To find out: ∠ADCIn △ADB and △ADC,

90 ∘ Given that, in △ABC,AB=AC, BD=DC To find out: ∠ADCIn △ADB and △ADC,AB=AC[given]

90 ∘ Given that, in △ABC,AB=AC, BD=DC To find out: ∠ADCIn △ADB and △ADC,AB=AC[given] BD=DC[given]

90 ∘ Given that, in △ABC,AB=AC, BD=DC To find out: ∠ADCIn △ADB and △ADC,AB=AC[given] BD=DC[given]Also, AD=AD[common]

90 ∘ Given that, in △ABC,AB=AC, BD=DC To find out: ∠ADCIn △ADB and △ADC,AB=AC[given] BD=DC[given]Also, AD=AD[common]∴ By SSS congruence criterion, ΔADB≅ΔADC.

90 ∘ Given that, in △ABC,AB=AC, BD=DC To find out: ∠ADCIn △ADB and △ADC,AB=AC[given] BD=DC[given]Also, AD=AD[common]∴ By SSS congruence criterion, ΔADB≅ΔADC.Hence, ∠ADB=∠ADB

90 ∘ Given that, in △ABC,AB=AC, BD=DC To find out: ∠ADCIn △ADB and △ADC,AB=AC[given] BD=DC[given]Also, AD=AD[common]∴ By SSS congruence criterion, ΔADB≅ΔADC.Hence, ∠ADB=∠ADBAlso, ∠ADB+∠ADC=180

90 ∘ Given that, in △ABC,AB=AC, BD=DC To find out: ∠ADCIn △ADB and △ADC,AB=AC[given] BD=DC[given]Also, AD=AD[common]∴ By SSS congruence criterion, ΔADB≅ΔADC.Hence, ∠ADB=∠ADBAlso, ∠ADB+∠ADC=180 ∘

90 ∘ Given that, in △ABC,AB=AC, BD=DC To find out: ∠ADCIn △ADB and △ADC,AB=AC[given] BD=DC[given]Also, AD=AD[common]∴ By SSS congruence criterion, ΔADB≅ΔADC.Hence, ∠ADB=∠ADBAlso, ∠ADB+∠ADC=180 ∘ [linear pair]

90 ∘ Given that, in △ABC,AB=AC, BD=DC To find out: ∠ADCIn △ADB and △ADC,AB=AC[given] BD=DC[given]Also, AD=AD[common]∴ By SSS congruence criterion, ΔADB≅ΔADC.Hence, ∠ADB=∠ADBAlso, ∠ADB+∠ADC=180 ∘ [linear pair]∴ ∠ADB=∠ADC=90

90 ∘ Given that, in △ABC,AB=AC, BD=DC To find out: ∠ADCIn △ADB and △ADC,AB=AC[given] BD=DC[given]Also, AD=AD[common]∴ By SSS congruence criterion, ΔADB≅ΔADC.Hence, ∠ADB=∠ADBAlso, ∠ADB+∠ADC=180 ∘ [linear pair]∴ ∠ADB=∠ADC=90 ∘

90 ∘ Given that, in △ABC,AB=AC, BD=DC To find out: ∠ADCIn △ADB and △ADC,AB=AC[given] BD=DC[given]Also, AD=AD[common]∴ By SSS congruence criterion, ΔADB≅ΔADC.Hence, ∠ADB=∠ADBAlso, ∠ADB+∠ADC=180 ∘ [linear pair]∴ ∠ADB=∠ADC=90 ∘

90 ∘ Given that, in △ABC,AB=AC, BD=DC To find out: ∠ADCIn △ADB and △ADC,AB=AC[given] BD=DC[given]Also, AD=AD[common]∴ By SSS congruence criterion, ΔADB≅ΔADC.Hence, ∠ADB=∠ADBAlso, ∠ADB+∠ADC=180 ∘ [linear pair]∴ ∠ADB=∠ADC=90 ∘ Hence, ∠ADC=90

90 ∘ Given that, in △ABC,AB=AC, BD=DC To find out: ∠ADCIn △ADB and △ADC,AB=AC[given] BD=DC[given]Also, AD=AD[common]∴ By SSS congruence criterion, ΔADB≅ΔADC.Hence, ∠ADB=∠ADBAlso, ∠ADB+∠ADC=180 ∘ [linear pair]∴ ∠ADB=∠ADC=90 ∘ Hence, ∠ADC=90 ∘

Answer 2

GIVEN:

In △ABC,

AB=AC, BD=DC

TO FIND: ∠ADC

In △ADB and △ADC,

AB=AC [given]

BD=DC [given]

Also, AD=AD [common]

∴ By SSS congruence criterion, ΔADB≅ΔADC.

Hence, ∠ADB=∠ADB

Also, ∠ADB+∠ADC=180∘

[linear pair]

∴ ∠ADB=∠ADC=90∘

HENCE:

∠ADC=90∘