MHB Prove Triangle BAD = CEA: Tips & Maths Class Help

[Mathick]

In the triangle $$ABC$$ a point $$D$$ lies on the edge $$BC$$, $$E$$ - on the edge $$AB$$. Aditionally, $$BD=AC$$, $$AD=AE$$ and $$AB^2=AC\cdot BC$$. Prove that $$\sphericalangle BAD = \sphericalangle CEA$$.

I have to do this task on Maths class on Monday and I need any tips or something because I don't know how to start it.

 

[Opalg]

In the triangle $$ABC$$ a point $$D$$ lies on the edge $$BC$$, $$E$$ - on the edge $$AB$$. Aditionally, $$BD=AC$$, $$AD=AE$$ and $$AB^2=AC\cdot BC$$. Prove that $$\sphericalangle BAD = \sphericalangle CEA$$.

I have to do this task on Maths class on Monday and I need any tips or something because I don't know how to start it.

View attachment 4729​

I won't give you the answer, but see if you can show that (1) the line $AB$ is tangent to the circle through $A$, $C$ and $D$; (2) the triangles $BAD$, $CEA$ are congruent.

 

[Mathick]

​

I won't give you the answer, but see if you can show that (1) the line $AB$ is tangent to the circle through $A$, $C$ and $D$; (2) the triangles $BAD$, $CEA$ are congruent.

Proof: the line $AB$ is tangent to the circle through $A$, $C$ and $D$

Knowing that $$BD=AC$$, $$AD=AE$$ and $$AB^2=AC\cdot BC$$, we get $$\frac{AB}{AC}=\frac{BC}{AB}$$ and then $$\frac{AB}{BD}=\frac{BC}{AB}$$. Thus, the triangles $$ABD$$ and $$CBA$$ are similar. So $$\sphericalangle BDA=\sphericalangle CAB$$ and $$\sphericalangle BAD=\sphericalangle ACB$$. Additionally, $$\frac{AC}{AB}=\frac{AD}{BD}$$ (because the triangles $$ABD$$ and $$CBA$$ are similar). So $$\sphericalangle BAC=\sphericalangle BDA$$, and what's more $$\sphericalangle ADC=\sphericalangle CA...$$ (there is no letter on line AB after A).

As a result, the line AB is tangent to the circle through $A$, $C$ and $D$ because of the theorem of which is illustrated on the photo.

View attachment 4735

Is it correct? Is there any better proof?