Possible criterion for proving parallelogram.

Question

A quadrilateral $ABCD$ has $∠BAD = ∠BCD$ and diagonal $AC$ bisects diagonal $BD$ at $P$. Is it necessarily a parallelogram? If not, give an example of such a quadrilateral. Provide proof.

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I was thinking about this, please tell if I am wrong.

Assume $∠BAC > ∠ACD$, for now.

Taking $∠BAC = ∠ACD + k$ and $∠BCA = ∠CAD + k$

Marking $∠BAX = k $ and $∠BCY = k$, where $X$ and $Y$ are points on $BD$.

proved $AX || BC$ and $CY || AD$

$∠CAD = ∠ACY$ which proves

$∆PAD$ ~ $∆PCY$

ie. $AP/PC = DP/PY$

(From this point on see user’s answer. The following part is WRONG)

And similarly, $∆CPD$ ~ $∆APX$

ie. $AP/PC = DP/PX$

and using these similarity ratios showed that $X = Y$.

Hence $AXCD$ is a parallelogram; where at $P$, $AP = PC,\,PX = DP = PB$ which proved $X = B$. So $ABCD $ must be a parallelogram.

Similarly, we can prove for $∠BAC < ∠ACD$, and for $∠BAC = ∠ACD$ it is obvious.

Please use highschool geometry.

Answer 1

Big hint: As the angles at $A$ and $C$ are equal both points must lie on arc of the same radius over $BD$.

Obviously -- due to symmetry -- triangles $ADB$ and $CBD$ are congruent.

Answer 2

You are wrong in the first line itself. If you want to take this approach then you have to use directed angles modulo $\pi$ or take separate boring cases, how do you know that $\angle BAC \geq \angle ACD$? Also, how are the two $k$'s the same in both the expressions?

Answer 3

Clearly you proof is false, as you have succeeded to prove $X=Y$ without using the equality $BP=PD$ and without this condition the claim is not valid as the next drawing (where $\angle BAD=\angle BCD$) shows:

There arises of course the question: what is wrong with your "simple similarity ratios"? Elaborating the issue one eventually finds the source (which was hidden behind the word "similarly"): $$ \triangle PAD\sim \triangle PCY\implies \frac{PY}{PC}=\frac{PD}{PA}\implies\frac{PY}{PD}=\frac{PC}{PA},\\ \triangle PCD\sim \triangle PAX\implies \frac{PX}{PA}=\frac{PD}{PC}\implies\frac{PX}{PD}=\color{red}{\frac{PA}{PC}}.\\ $$

This invalidates the whole "proof" (assuming $PA\ne PC$).

Answer 4

(I did not look at your proof.)

Draw the circumcircle $\gamma$ of the triangle $ABD$. Let $\gamma'$ be the circle obtained by reflecting $\gamma$ on the line $B\vee D$. The two circles then form a fourfold symmetric part of the figure, because there is also a symmetry with respect to the normal of $B\vee D$ erected at $P$.

We have $C\in \gamma'$ because of the peripheral angles, and $C\in A\vee P$. It follows that $|CP|=|AP|$, because of the fourfold symmetry. If $\sigma$ is the point-reflection of the plane at $P$ then $$\sigma:\quad\triangle(ADC)\to\triangle(CBA)\ .$$ This implies that the sides of $ABCD$ are pairwise parallel.