the bisertor AD of angle BAC of triangle ABC bisect as the side BD at D . prove that Triangle ABC is an isosceles triangle
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In ΔABC, let the altitude AD bisects ∠BAC
In ΔABC, let the altitude AD bisects ∠BACThen we have to prove that the ΔABC is isosceles.
In ΔABC, let the altitude AD bisects ∠BACThen we have to prove that the ΔABC is isosceles.In triangle ADB and ADC,
In ΔABC, let the altitude AD bisects ∠BACThen we have to prove that the ΔABC is isosceles.In triangle ADB and ADC,∠BAD=∠CAD (AD is bisector of ∠BAC)
In ΔABC, let the altitude AD bisects ∠BACThen we have to prove that the ΔABC is isosceles.In triangle ADB and ADC,∠BAD=∠CAD (AD is bisector of ∠BAC)AD=AD (common)
In ΔABC, let the altitude AD bisects ∠BACThen we have to prove that the ΔABC is isosceles.In triangle ADB and ADC,∠BAD=∠CAD (AD is bisector of ∠BAC)AD=AD (common)∠ADB=∠ADC (Each equal to 90
In ΔABC, let the altitude AD bisects ∠BACThen we have to prove that the ΔABC is isosceles.In triangle ADB and ADC,∠BAD=∠CAD (AD is bisector of ∠BAC)AD=AD (common)∠ADB=∠ADC (Each equal to 90 o
In ΔABC, let the altitude AD bisects ∠BACThen we have to prove that the ΔABC is isosceles.In triangle ADB and ADC,∠BAD=∠CAD (AD is bisector of ∠BAC)AD=AD (common)∠ADB=∠ADC (Each equal to 90 o )
In ΔABC, let the altitude AD bisects ∠BACThen we have to prove that the ΔABC is isosceles.In triangle ADB and ADC,∠BAD=∠CAD (AD is bisector of ∠BAC)AD=AD (common)∠ADB=∠ADC (Each equal to 90 o )⇒ΔADB≅ΔADC (by ASA congruence criterion)
In ΔABC, let the altitude AD bisects ∠BACThen we have to prove that the ΔABC is isosceles.In triangle ADB and ADC,∠BAD=∠CAD (AD is bisector of ∠BAC)AD=AD (common)∠ADB=∠ADC (Each equal to 90 o )⇒ΔADB≅ΔADC (by ASA congruence criterion)⇒AB=AC (cpct)
In ΔABC, let the altitude AD bisects ∠BACThen we have to prove that the ΔABC is isosceles.In triangle ADB and ADC,∠BAD=∠CAD (AD is bisector of ∠BAC)AD=AD (common)∠ADB=∠ADC (Each equal to 90 o )⇒ΔADB≅ΔADC (by ASA congruence criterion)⇒AB=AC (cpct)Hence, ΔABCis an isosceles.
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