Fig. 1.629In the figure, in ABC. point D onSide BC is such that.angle

1.	Fig. 1.629In the figure, in ABC. point D onSide BC is such that.angle BAC=angle ADCProve that, CA^2= CB XCD
Answer» ANSWER: Hi! hope this HELPS:) brainiest answer? Step-by-step explanation: GIVEN in ΔABC, ∠ADC = ∠BAC Consider ΔBAC and ΔADC ∠ADC = ∠BAC (Given) ∠C = ∠C (Common angle) ∴ ΔBAC ~ ΔADC (AA similarity CRITERION) AB/AD=CB/CA=CA/CD Consider CB/CA=CA/CD CA^2=CB×CD

Fig. 1.629In the figure, in ABC. point D onSide BC is such that.angle BAC=angle ADCProve that, CA^2= CB XCD

Answer»

ANSWER:

Hi! hope this HELPS:) brainiest answer?

Step-by-step explanation:

GIVEN in ΔABC, ∠ADC = ∠BAC

Consider

ΔBAC and ΔADC

∠ADC = ∠BAC (Given)

∠C = ∠C (Common angle)

∴ ΔBAC ~ ΔADC (AA similarity CRITERION)

AB/AD=CB/CA=CA/CD

Consider CB/CA=CA/CD

CA^2=CB×CD

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Fig. 1.629In the figure, in ABC. point D onSide BC is such that.angle BAC=angle ADCProve that, CA^2= CB XCD​

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Fig. 1.629In the figure, in ABC. point D onSide BC is such that.angle BAC=angle ADCProve that, CA^2= CB XCD