If angle of a triangle to the opposite side are equal than prove that triangle is isosceles.
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Let the triangle be ABC
Angle B=angle C (according to the question )
Draw an altitude from A to D on BC
In ∆ABD and ∆ACD
Angle B=angle C (given)
Angle ADB = angle ADC = 90°(by construction)
AD= AD (common)
By AAS criteria ∆ABD and ∆ACD are congruent
AB= AC (cpct) Thus ∆ABC is an isosceles triangle
Angle B=angle C (according to the question )
Draw an altitude from A to D on BC
In ∆ABD and ∆ACD
Angle B=angle C (given)
Angle ADB = angle ADC = 90°(by construction)
AD= AD (common)
By AAS criteria ∆ABD and ∆ACD are congruent
AB= AC (cpct) Thus ∆ABC is an isosceles triangle
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