# Proving two triangles are congruent The two circles shown are identical and pass through each other's centre. The line $$AC$$ passes through the centres of both circles.

I would like to prove that triangles $$ABD$$ and $$BCE$$ are congruent (I think they are?)

I want the shortest possible reasoning.

Currently, I think I have it by side-side-angle.

Clearly, $$AD=EC$$ and $$BD=BE$$ because they are radii of the same circle.

$$BDE$$ is an isosceles triangle with angles $$BDE$$ and $$BED$$ being equal (call this value $$x$$). But since $$ADE$$ is a straight line, as is $$DEC$$, we have $$BDA = 180 - x$$ and $$BEC = 180 - x$$.

Hence the triangles are congruent by side-side-angle.

IS this correcT?

Are there any alternatives? Would be interested in as many as possible - in particular, can we argue $$BC =AB$$ through obvious means or any of the other angles are equal? :)

• Note that triangle $BDE$ is equilateral (each side is a radius). – quasi Aug 17 at 21:23
• Side-side-angle for congruence is incorrect. However you have side-angle-side which is correct to show congruence. – herb steinberg Aug 17 at 22:17
• "Hence the triangles are congruent by side-side-angle."" There's no such thing as congruent by side-side-angle. If you are going to prove two triangle are congruent by two sides and an angle the angle must be between the two sides to get side-angle-side (which is what you do have). If you have angle-side-side that will not be enough to prove they are congruent and you will end up making an A.S.S. of yourself. – fleablood Aug 17 at 23:16
• @fleablood nice one, I didn’t actually appreciate the difference between side side angle and side angle side!! – PhysicsMathsLove Aug 17 at 23:18
• but note you have sides $AD\cong CE$ and $\angle ADB\cong CEB$ and $DB = EB$. So you have S.A.S. which is good. You do not have S.S.A which would have been bad. – fleablood Aug 17 at 23:20

Unfortunately, side side angle is not a valid congruence (see if you can figure out why). But in your case you have side angle side, which is valid. One easier, maybe more intuitive way to see that $$BC=AB$$, is to understand that the figure is symmetrical about the line through the two intersection points of the circle. Although this way may not be the easiest to rigorously prove, it is one that I would say makes the most sense. Alternative proofs can be: proving $$△ABE \cong △CBD$$, $$\angle A \cong \angle C$$ by arcs, or, if you're feeling like going a different route, you can prove that since $$CB$$ and $$AB$$ are tangents of equal distance to congruent circles, they must be the same length.
You can easily show that $$\triangle ABE \cong \triangle BDC$$ because both are right triangles (inscribed angles subtended by a diameter are right) and have same hypotenuse and one of the legs.
If $$BD$$and $$BE$$ are congruent, then the triangle containing these sides is isosceles with its base angles at $$D$$and $$E$$ congruent. Then the supplements of the latter two angles are also congruent, and the claimed triangle congruence is proved by SAS.