Question

prove that the angle opposite to equal sides of a triangle are equal ?
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Answer 1

ABC is a given triangle with, AB=AC.

ABC is a given triangle with, AB=AC.To prove: Angle opposite to AB= Angle

ABC is a given triangle with, AB=AC.To prove: Angle opposite to AB= Angleopposite to AC (i.e) ∠C=∠B

ABC is a given triangle with, AB=AC.To prove: Angle opposite to AB= Angleopposite to AC (i.e) ∠C=∠BConstruction: Draw AD perpendicular to BC

ABC is a given triangle with, AB=AC.To prove: Angle opposite to AB= Angleopposite to AC (i.e) ∠C=∠BConstruction: Draw AD perpendicular to BC∴∠ADB=∠ADC=90

ABC is a given triangle with, AB=AC.To prove: Angle opposite to AB= Angleopposite to AC (i.e) ∠C=∠BConstruction: Draw AD perpendicular to BC∴∠ADB=∠ADC=90 o

ABC is a given triangle with, AB=AC.To prove: Angle opposite to AB= Angleopposite to AC (i.e) ∠C=∠BConstruction: Draw AD perpendicular to BC∴∠ADB=∠ADC=90 oProof:

ABC is a given triangle with, AB=AC.To prove: Angle opposite to AB= Angleopposite to AC (i.e) ∠C=∠BConstruction: Draw AD perpendicular to BC∴∠ADB=∠ADC=90 oProof: Consider △ABD and △ACD

ABC is a given triangle with, AB=AC.To prove: Angle opposite to AB= Angleopposite to AC (i.e) ∠C=∠BConstruction: Draw AD perpendicular to BC∴∠ADB=∠ADC=90 oProof: Consider △ABD and △ACDAD is common

ABC is a given triangle with, AB=AC.To prove: Angle opposite to AB= Angleopposite to AC (i.e) ∠C=∠BConstruction: Draw AD perpendicular to BC∴∠ADB=∠ADC=90 oProof: Consider △ABD and △ACDAD is commonAB=AC

ABC is a given triangle with, AB=AC.To prove: Angle opposite to AB= Angleopposite to AC (i.e) ∠C=∠BConstruction: Draw AD perpendicular to BC∴∠ADB=∠ADC=90 oProof: Consider △ABD and △ACDAD is commonAB=AC∴∠ADB=∠ADC=90

ABC is a given triangle with, AB=AC.To prove: Angle opposite to AB= Angleopposite to AC (i.e) ∠C=∠BConstruction: Draw AD perpendicular to BC∴∠ADB=∠ADC=90 oProof: Consider △ABD and △ACDAD is commonAB=AC∴∠ADB=∠ADC=90 o

ABC is a given triangle with, AB=AC.To prove: Angle opposite to AB= Angleopposite to AC (i.e) ∠C=∠BConstruction: Draw AD perpendicular to BC∴∠ADB=∠ADC=90 oProof: Consider △ABD and △ACDAD is commonAB=AC∴∠ADB=∠ADC=90 oHence ∠ABD=∠ACD

ABC is a given triangle with, AB=AC.To prove: Angle opposite to AB= Angleopposite to AC (i.e) ∠C=∠BConstruction: Draw AD perpendicular to BC∴∠ADB=∠ADC=90 oProof: Consider △ABD and △ACDAD is commonAB=AC∴∠ADB=∠ADC=90 oHence ∠ABD=∠ACD∠ABC=∠ACB

ABC is a given triangle with, AB=AC.To prove: Angle opposite to AB= Angleopposite to AC (i.e) ∠C=∠BConstruction: Draw AD perpendicular to BC∴∠ADB=∠ADC=90 oProof: Consider △ABD and △ACDAD is commonAB=AC∴∠ADB=∠ADC=90 oHence ∠ABD=∠ACD∠ABC=∠ACB∠B=∠C. Hence the proof

ABC is a given triangle with, AB=AC.To prove: Angle opposite to AB= Angleopposite to AC (i.e) ∠C=∠BConstruction: Draw AD perpendicular to BC∴∠ADB=∠ADC=90 oProof: Consider △ABD and △ACDAD is commonAB=AC∴∠ADB=∠ADC=90 oHence ∠ABD=∠ACD∠ABC=∠ACB∠B=∠C. Hence the proofThis is known as Isosceles triangle theorem