If the bisector of exterior vertical angle of a triangle is parallel to the base , show that the triangle is isosceles
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Hey mate..
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∠CAE is the external vertical angle and AD is its bisector.
∴ ∠CAD = ∠DAE ...... (i)
As, AD || BC
∴ ∠CAD = ∠ACB ...... (ii) (alternate angle)
and ∠DAE = ∠ABC ......(iii) (corresponding angle)
From (i), (ii) and (iii),
∠ABC = ∠ACB
∴ AC = AB (opposite sides to equal angles)
Hence, ΔABC is isosceles.
Hope it helps !
========
∠CAE is the external vertical angle and AD is its bisector.
∴ ∠CAD = ∠DAE ...... (i)
As, AD || BC
∴ ∠CAD = ∠ACB ...... (ii) (alternate angle)
and ∠DAE = ∠ABC ......(iii) (corresponding angle)
From (i), (ii) and (iii),
∠ABC = ∠ACB
∴ AC = AB (opposite sides to equal angles)
Hence, ΔABC is isosceles.
Hope it helps !
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