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Isosceles Triangle Median Diagram Proof

Organize and draw diagrams for problems concerning isosceles triangles and their medians. Prove that the median is the perpendicular bisector of the base.

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Isosceles Triangle Median Diagram Proof

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  1. Warm Up

  2. Chapter 4.2 The Case of the Missing Diagram

  3. Organize the information in, and draw diagrams for, problems presented in words.

  4. Set up this problem: An isosceles triangle and the median to the base. Draw the shape, label everything Write the givens and what you want to prove.

  5. Given: an isosceles triangle and the median to the base. • Prove: The median is the perpendicular bisector of the base. • Notice: There are two conclusions to made: • The median is perpendicular to the base. • The median bisects the base.

  6. A C B D Now draw and label all you know. You can label everything on the diagram to help you make the proof. Given: ABC is isosceles Base BC AD is a median Prove: BD AD and bisects BC

  7. NOTICE!!! You can label everything on a diagram to help you make the proof. Some problems you only have to draw, label, write the givens and what to prove. Others you also have to prove.

  8. Remember If….then…. Sometimes you will see these in reverse. The medians of a triangle are congruent if the triangle is equilateral. Draw and set up the proof. Write down the givens you need. What do you need to prove?

  9. X Given: Δ XYZ is equilateral PY, RZ and QX are medians Prove: PY = RZ = QX P R ~ ~ Y Z Q

  10. The median to the base of an isosceles triangle divides the triangle into two congruent triangles. Draw, write givens and what to prove, then prove.

  11. C T A R Given: Δ CAT is isosceles, with base TA. CR is a median. Prove: Δ TRC = Δ ARC ~

  12. Try this one! If each pair of opposite sides of a four-sided figure are congruent, then the segments joining opposite vertices bisect each other. Draw Write Given: Write Prove: Write proof!

  13. A B C D E ~ Given: AB = CD AD = BC Prove: AC bisects BD BD bisects AC ~

  14. A B C D ~ Δ ABC = Δ CDA by SSS, and thus, <BAC = <DCA. Δ BAD = Δ DCB by SSS, and thus, <ABD = <CDB. Thus Δ ABE = Δ CDE by ASA, and then AE = EC and DE = EB. ~ ~ ~ ~ ~ ~ E

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