Prove that: “ Angles opposite to equal sides of an isosceles triangle are equal”
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Step-by-step explanation:
ABC is a given triangle with, AB=AC.
To prove: Angle opposite to AB= Angle
opposite to AC (i.e) ∠C=∠B
Construction: Draw AD perpendicular to BC
∴∠ADB=∠ADC=90
o
Proof:
Consider △ABD and △ACD
AD is common
AB=AC
∴∠ADB=∠ADC=90
o
Hence ∠ABD=∠ACD
∠ABC=∠ACB
∠B=∠C. Hence the proof
This is known as Isosceles triangle theorem
solution
Answered by
0
Step-by-step explanation:
ABC is a given triangle with, AB=AC.
To prove: Angle opposite to AB= Angle
opposite to AC (i.e) ∠C=∠B
Construction: Draw AD perpendicular to BC
∴∠ADB=∠ADC=90
Proof:
Consider △ABD and △ACD
AD is common
AB=AC
∴∠ADB=∠ADC=90
Hence ∠ABD=∠ACD
∠ABC=∠ACB
∠B=∠C. Hence proof
This is known as Isosceles triangle theorem
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