Show that the bisector of the vertical angle of an isosceles triangle bisect the base at right angles
use the law of sines to show that the base is divided into equal lengths.
Then that will imply that the angles at the base of the altitude are also equal, so they are both 90 degrees.
Show that the bisector of the vertical of an isosceles triangle bisects the base at tight angles
To show that the bisector of the vertical angle of an isosceles triangle bisects the base at right angles, we can use the properties of triangles and angle bisectors.
Let's consider an isosceles triangle △ABC, where AB = AC. We want to prove that the bisector of the vertical angle at A bisects the base BC at right angles.
1. Draw the isosceles triangle: Start by drawing an isosceles triangle with AB = AC.
2. Draw the angle bisector: Draw the bisector of the angle BAC, starting from point A.
3. Label the intersection point: Let the bisector intersect the base BC at point D.
4. Prove AD = DC: To show that AD = DC, we can use the fact that the bisector divides the vertical angle at A into two congruent angles. This means that ∠DAB = ∠DAC.
5. Prove △ABD ≅ △ACD: Now, we can prove that triangles △ABD and △ACD are congruent. We already have AD = DC (proved in step 4), AB = AC (isosceles triangle), and ∠DAB = ∠DAC (angle bisector). By the side-angle-side (SAS) congruence postulate, we can conclude that △ABD ≅ △ACD.
6. Prove ∠ADB = ∠ADC: Since △ABD ≅ △ACD, the corresponding angles are also congruent. Therefore, ∠ADB = ∠ADC.
7. Prove ∠ADB + ∠ADC = 180°: The sum of the angles in a triangle is always 180°. Since ∠ADB = ∠ADC, we can write the equation as: ∠ADB + ∠ADB = 180°.
8. Solve the equation: Combining like terms, we get: 2∠ADB = 180°. Dividing both sides by 2, we have: ∠ADB = 90°.
9. Conclusion: We have shown that ∠ADB = 90°, which means that the bisector of the vertical angle at A bisects the base BC at right angles.
Therefore, the bisector of the vertical angle of an isosceles triangle does indeed bisect the base at right angles.