SOLUTION: PROVE: If an isosceles triangle has an altitude from the vertex to the base, then the altitude bisects the vertex angle. GIVEN: triangle ABC is isosceles; line CD is the altitude

Question 159413: PROVE:
If an isosceles triangle has an altitude from the vertex to the base, then the altitude bisects the vertex angle.
GIVEN: triangle ABC is isosceles; line CD is the altitude to base of line AB
TO PROVE: line CD bisects angle ACB
Problem says that I have to come up with a plan of the proof as well.
HEEELP!!
TIA, Joanne
Found 2 solutions by vleith, gonzo:
Answer by vleith(2983) (Show Source): You can put this solution on YOUR website!
Draw the isosceles triangle. Then draw the altitude.
That is what you are "given"
You also know that two angles of the isosceles triangle are equal. Let the two equal angles in the original isosceles triangle be A.
Euclid tells us that every triangle has 180 degrees of interior angle in it.
So the third angle in the isosceles triangle is (180-2A)
By definition, an altitude forms a right angle with the base it intersects.
Thus there are 2 right angles formed at the foot of the altitude.
So, you can now show that the two angles formed at the vertex where the altitude was dropped from must be equal (each being 180-90-A) = 90-A
(90-A) = 1/2(180-2A)
Thus the two new angles are equal, and are 1/2 the size of the original one. So the original angle is bisected by the altitude.

Answer by gonzo(654) (Show Source): You can put this solution on YOUR website!
your plan is as follows:
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prove that CD is perpendicular to AB.
prove that angles opposite congruent sides of isosceles triangle are congruent. assume this is previously proven. if not, you'll have to prove it yourself. i'll provide that proof separately.
prove that the two right triangles created by the altitude are congruent.
prove that angles created by altitude are equal to each other.
prove that they bisect the angle.
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proof that angles opposite congruent sides of isosceles triangle are congruent
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triangle ABC is isosceles and AC = CB (given)
draw CD to intersect AB so that AD = DB (construction)
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)
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your main proof.
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CD is perpendicular to AD (this is by definition of the altitude of a triangle)
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angle CAD congruent to angle CBD (opposite angles of an isosceles triangle are equal). just stating it should be enough but if they want proof, it is up above.
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triangle CDA is congruent to CDB (SSA - two triangle are congruent if two corresponding sides and an angle not between them are congruent. this is a basic postulate of congruent triangles. corresponding sides are CA congruent to CB, CD congruent to CD, corresponding angles are CAD congruent to CBD)
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angle ACD = angle BCD (corresponding angles of congruent triangles are congruent)
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CD bisects angle ACB (by definition the bisector of an angle creates two equal angles formed by the bisector)

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