Show that in an isosceles triangle, the angles opposite the equal sides are equal.(hint: Draw a perpendicular to the unequal side from the vertex)

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Solution

Given- $A B = A C$
To prove- $∠ B = ∠ C$
Construction-Draw $A D ⊥ B C$
Proof-In triangles ADB and ADC
$A B = A C$ (given)
$A D = A D$ (common)
$∠ A D B = ∠ A D C$ $(90^{o})$
Then by RHS criteria we have triangle ADB $≅$ triangle ADC
Hence, by $c . p . c . t$ $∠ B = ∠ C$

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Show that in an isosceles triangle, the angles opposite the equal sides are equal.(hint: Draw a perpendicular to the unequal side from the vertex)

Given-AB=ACTo prove-∠B=∠CConstruction-Draw AD⊥BCProof-In triangles ADB and ADCAB=AC (given)AD=AD (common)∠ADB=∠ADC (90o)Then by RHS criteria we have triangle ADB ≅ triangle ADCHence, by c.p.c.t ∠B=∠C

Given- $A B = A C$
To prove- $∠ B = ∠ C$
Construction-Draw $A D ⊥ B C$
Proof-In triangles ADB and ADC
$A B = A C$ (given)
$A D = A D$ (common)
$∠ A D B = ∠ A D C$ $(90^{o})$
Then by RHS criteria we have triangle ADB $≅$ triangle ADC
Hence, by $c . p . c . t$ $∠ B = ∠ C$