answer the question in the image
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i) IN TRIANGLE ADB AND ADC
AD=AD (COMMON)
AB=AC (GIVEN)
BD=DC (AD IS MEDIAN WHICH CUT TRIANGLE IN TWO EQUAL PART)
BY SSS
/\ ADB congruent /\ ADC
ii) ANGLE B = ANGLE C
BY C.P.C.T
AD=AD (COMMON)
AB=AC (GIVEN)
BD=DC (AD IS MEDIAN WHICH CUT TRIANGLE IN TWO EQUAL PART)
BY SSS
/\ ADB congruent /\ ADC
ii) ANGLE B = ANGLE C
BY C.P.C.T
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(i) In ΔADB & ΔADC
AB = AC [Given]
∠BAD = ∠DAC [AD is the bisector of ∠BAC]
AD = AD [ Common side]
∴ ΔADB ≅ ΔADC [ Side Angle Side Criterion]
(ii) Since, ΔADB ≅ ΔADC
∠B = ∠C [ corresponding part of congruent triangles(C.P.C.T) ]
Another Method:
In ΔABC
AB = AC
BC is the common base
∴∠B = ∠C (proved)
I hope it helps you ^_^
AB = AC [Given]
∠BAD = ∠DAC [AD is the bisector of ∠BAC]
AD = AD [ Common side]
∴ ΔADB ≅ ΔADC [ Side Angle Side Criterion]
(ii) Since, ΔADB ≅ ΔADC
∠B = ∠C [ corresponding part of congruent triangles(C.P.C.T) ]
Another Method:
In ΔABC
AB = AC
BC is the common base
∴∠B = ∠C (proved)
I hope it helps you ^_^
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