$△$ ABC is an isosceles triangle, in which AB=AC. Side BA is produced to D such that AD=AB. $∠$ BCD is equal to ___.

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Solution

Given, AB=AC and BA=AD

In $△$ ABC,

$∠$ ABC = $∠$ ACB = x (opposite angles of equal sides)

In $△$ ACD,

$∠$ ADC = $∠$ ACD = y (opposite angles of equal sides)

In $△$ BCD,

$∠$ DBC + $∠$ BCD + $∠$ CDB= $180^{\circ}$

x + x + y +y = $180^{\circ}$

2x+2y = $180^{\circ}$

x+y = $90^{\circ}$

$∠$ BCD = (x+y) = $90^{\circ}$

$∠$ BCD = $90^{\circ}$

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△ABC is an isosceles triangle, in which AB=AC. Side BA is produced to D such that AD=AB. ∠BCD is equal to ___.

Given, AB=AC and BA=AD In △ABC, ∠ABC = ∠ACB = x (opposite angles of equal sides) In △ACD, ∠ADC = ∠ACD = y (opposite angles of equal sides) In △BCD, ∠DBC + ∠BCD + ∠CDB= 180∘ x + x + y +y = 180∘ 2x+2y = 180∘ x+y = 90∘ ∠BCD = (x+y) = 90∘ ∠BCD = 90∘

$△$ ABC is an isosceles triangle, in which AB=AC. Side BA is produced to D such that AD=AB. $∠$ BCD is equal to ___.