The following is an image of a math problem I am trying to solve. It was originally a textbook problem (given O, P, Q are midpoints of CD, CB, and DA respectively and CM=CN and ∠COA=∠DOB, prove AD=CB, a trivial SAS congruence problem.) I changed the information into the following:
Given
- P is the midpoint of AO
- Q is the midpoint of BO
- O is the midpoint of CD
- CP = QD
- ∠COA = ∠DOB
Prove
- BC = AD
- AO = OB
What I have done
I have tried to prove ∆AOD is congruent to ∆BOC, but that is unfortunately SSA, which is famous for not being true.
I have succeeded in proving AB//PQ, but don't know what to do next.
Our class has only studied parallel lines and congruence and the sum of interior angles of a triangle being 180 degrees so if it is possible please do not use similarity or something clearly out of scope (for instance, trigonometry or the Cartesian Plane). Or, please tell me why I can't prove this. PAA (Partial Answers Appreciated).
Not sure whether you know Menelaus's theorem (the spelling is copied from wiki, which I find wierd).
Applying Menelaus to the triangle $AOD$ and the line $BC$, we get $\frac{AX}{XD} = \frac 12$. Similarly, $\frac{BX}{XC} = \frac 12$. These tell us that $AB$ is parallel to $CD$.
On the other hand, $\frac{OP}{PA} = 1 = \frac{OQ}{QB}$ which tells us that $PQ$ is parallel to $AB$.
Hence $PQ$ is parallel to $CD$. Together with $CP = QD$, we see that $CPQD$ is an isosceles trapezoid and everything else is clear.