Prove two triangles congruent by a side, a median drawn to that side, and the angle formed by the side and the median.

Let's call one triangle ABC, and M the midpoint of AB.

Let's call the other triangle DEF, and midpoint N of DE.

We have a congruent side, AB=DE, congruent medians CM=FN, and congruent angles, ∠AMC=∠DNF

AB ≅ DE  Given

AM ≅ BM and DN ≅ EN    Def median/midpoint

AM+BM=AB  and DN+EN=DE     Segment addition theorem

2AM=2BM=AB and 2DN=2EN=DE   Substitution (eg BM for AM)

AM ≅ BM ≅ DN ≅ EN        Transitivity, algebra

CM ≅ FN   Given

∠AMC ≅ ∠DNF   Given

ΔAMC ≅ ΔDNF     Side Angle Side

AC ≅  DF   CPCTC

∠BMC ≅ ∠ENF   Supplements of congruent angles are congruent

ΔBMC ≅ ΔENF     Side Angle Side

BC ≅  EF      CPCTC

Δ ABC ≅ Δ DEF   Side Side Side

step-by-step explanation:

x^2 -12x + 36 + y^2 = 36

x^2 + y^2 = 12x

r^2 = 12rcos(theta)

r = 12 cos(theta)