Question

prove that angles opposite to equal sides are equal in an isosceles triangle

Answer 1

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Answer 2

$\textbf{ \huge{ Hello Mate! }}$

$\textbf{ Given }$ : AC = BC

$\textbf{ To Construct }$ :CD perpendicular to AB

$\textbf{ To prove }$ : <A = <B

$\textbf{ Proof }$ : In triangles ADC and BDC we get,

CD = CD ( Common )

<ADC = <BDC ( 90° each )

AC = BC ( Given )

Hence, by RHS congruency both triangles are congruent.

< A = < B ( c.p.c.t )

$\textbf{ \large{ Q.E.D }}$

Have great future ahead!