this is a basic postulate of congruent triangles. ![]() Let the vectors from the center of the circumscribed circle to the vertices of the triangle be A, B and C. Now for simplicity we will use different notation from the first proof. Triangle CDA is congruent to CDB (SSA - two triangle are congruent if two corresponding sides and an angle not between them are congruent. We show this using the dot product: ( U + V) ( U V) U 2 U V + V U V 2 0. ![]() just stating it should be enough but if they want proof, it is up above. Triangle ADC congruent to triangle BDC by SSS (AC = CB is given, CD = CD by reflexive property of equality (anything is equal to itself), AD = DB by construction)ĬD is perpendicular to AD (this is by definition of the altitude of a triangle)Īngle CAD congruent to angle CBD (opposite angles of an isosceles triangle are equal). Triangle ABC is isosceles and AC = CB (given)ĭraw CD to intersect AB so that AD = DB (construction) Proof that angles opposite congruent sides of isosceles triangle are congruent If the altitude is drawn to the hypotenuse of a right triangle, then the two triangles formed are similar to the original. Prove that angles created by altitude are equal to each other. And although the theorem can be proven in many different ways, many of the proofs are at least partially algebraic, and therefore do not provide an intuitive, geometric sense why it is true. Prove that the two right triangles created by the altitude are congruent. if not, you'll have to prove it yourself. Prove that angles opposite congruent sides of isosceles triangle are congruent. You can put this solution on YOUR website!
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