Math, asked by dubenikhil331, 2 months ago

. In Fig. 14.172, DE is a tangent to the circumcircle of
A ABC at the vertex A such that DE || BC. Show that
AB = AC

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Answered by riservikrant

Step-by-step explanation:

$given \: that \\ < eac = < acb \\ <dab = < abc \\ from \: linear \: pair \: angle \\ <dab + < bac + < eac = 180 \\ <abc + < bac + < acb = 180 \: (1 {eq}^{n} ) \\ in \: triangle \: abc \\ <bca + < bac < + < abc = 180 \\ from \: (1 \: {eq}^{n} ) \\ 180 - bac + < bac = 180 \\ <bac = < bca \\ < a = <c \: \: { proved}$

triangle abc is isoscales triangle

therefore, AB =AC

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. In Fig. 14.172, DE is a tangent to the circumcircle ofA ABC at the vertex A such that DE || BC. Show thatAB = AC​

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. In Fig. 14.172, DE is a tangent to the circumcircle of
A ABC at the vertex A such that DE || BC. Show that
AB = AC