$A,B,C,D$ on a circle. $\widehat{BAC}=\widehat{BDC}$

Question

Points $A,B,C,D$ belong to a circle. What's a rigorous yet simple proof that $\widehat{BAC}=\widehat{BDC}$ ?

Does this property have a name?

I get that in the above figure:

summing angles, $\widehat{BOA}+\widehat{AOC}=\widehat{BOC}=\widehat{BOD}+\widehat{DOC}$
sum of the angles of isosceles triangle $AOB$ is $\Pi$, thus
$\widehat{BOA}=\Pi-2\widehat{BAO}\quad$ and similarly
$\widehat{AOC}=\Pi-2\widehat{OAC}\quad$
$\widehat{BOD}=\Pi-2\widehat{BDO}\quad$
$\widehat{DOC}=\Pi-2\widehat{ODC}\quad$
replacing then simplifying, we get
$\widehat{BAO}+\widehat{OAC}=\widehat{BDO}+\widehat{ODC}\quad$ thus
$\widehat{BAC}=\widehat{BDC}\quad$ _^Q.E.D.

However this reasoning seems dependent on the order of points on the circle, and perhaps other hypothesis.

Vsauce's Michael (on the channel DONG) made this video about Thale's theorem recently. It's the special case where $BC$ is a diagonal. However, the ideas of the proof are applicable in general. You could take a look. — Arthur, Feb 04 '19 at 07:30
@abc... Nice pick. I (now) see that's also the Opera House theorem. — fgrieu, Feb 04 '19 at 07:36
@fgrieu: It's most-commonly known as (a corollary to) the Inscribed Angle Theorem. ("Opera House Theorem" may be local to Australia (and/or the author of that note), due to the figure's resemblance to the Sydney Opera House.) — Blue, Feb 04 '19 at 08:03

score 1 · Answer 1 · answered Feb 04 '19 at 07:22

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This is a well known property of angles, inscribed in a chord of a circle: they are congruent if they are inscribed on the same side of the chord (BC in your case) and supplementary if inscribed on opposite sides of the chord.

answered Feb 04 '19 at 07:22

GReyes

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Yes I (now) see that what I ask is a most direct application of that fact/theorem, which vaguely rings a bell. Yet that theorem is not in the curriculum of my daughter (11th grade) and we are seeking a rigorous simple proof of an exercise she's given, if there is one. – fgrieu Feb 04 '19 at 07:27

user638473 · Accepted Answer · 2019-02-06T16:48:34.740

Here is a proof that an angle inscribed in an arc is half the intersepted arc. That is, $m\angle APB=\frac{1}{2}m(arc \widehat{ACB})$

Following the figure, let $\mathrm{M}$ be the center of the circle, and $\angle APB=\alpha , \angle APM=\beta\\ \mathrm{Now, \: in \: \Delta MPB,}\\ \mathrm{MP}=\mathrm{MB}\\ \implies \angle MPB= \angle MBP= \alpha + \beta\\ \implies \angle CMB= 2 ( \alpha +\beta)\\ \: \\ \mathrm{In \: \Delta MPA,}\\ \mathrm{MP} =\mathrm{MA} \\ \angle MPA= \angle MAP= \beta \\ \implies \angle CMA=2\beta\\ \mathrm{Since \:} \angle CMB= \angle CMA+ \angle AMB\\ 2(\alpha+ \beta)= 2\beta + \angle AMB\\ \implies 2\alpha=\angle AMB\\ \: \\ \mathrm{And, \: since\:} \angle APB=\alpha, \mathrm{we \: have \: shown \:that} \\ \boxed{\angle APB=\frac{1}{2} \angle AMB} $

Now, your statement, that is, angles inscribed in the same arc are congruent, is a direct consequence of the previous theorem.

For in your case, $\angle BAC= \frac{1}{2}\angle BOC\\ \angle BDC=\frac{1}{2}\angle BOC\\ \implies \angle BAC=\angle BDC$

$A,B,C,D$ on a circle. $\widehat{BAC}=\widehat{BDC}$

2 Answers2