Find the measure of angle BAC.

[anemone]

Hi members of the forum,

Problem:
In triangle ABC, D is the midpoint of AB and E is the point of trisection of BC nearer to C. Given that $\displaystyle \angle ADC=\angle BAE$, find $\displaystyle\angle BAC$.

I have tried to solve it using only one approach, that is by assigning $\displaystyle\theta$ and $\displaystyle\beta$ to represent the angles of CAE and ABC respectively and applied the rules of Sine and Cosine appropriately but ended up with very messy equation with many unknown values.

View attachment 630

It's not that I didn't try it but now I feel like I've reached a plateau and I decided to ask for help at MHB.

If anyone could give me some hints, that would be great.

Thanks in advance.

 

[caffeinemachine]

Re: Find the angle of BAC.

Hi members of the forum,

Problem:
In triangle ABC, D is the midpoint of AB and E is the point of trisection of BC nearer to C. Given that $\displaystyle \angle ADC=\angle BAE$, find $\displaystyle\angle BAC$.

I have tried to solve it using only one approach, that is by assigning $\displaystyle\theta$ and $\displaystyle\beta$ to represent the angles of CAE and ABC respectively and applied the rules of Sine and Cosine appropriately but ended up with very messy equation with many unknown values.

https://www.physicsforums.com/attachments/630

It's not that I didn't try it but now I feel like I've reached a plateau and I decided to ask for help at MHB.

If anyone could give me some hints, that would be great.

Thanks in advance.

Let the point of trisection of $BC$ nearer to $B$ be $F$. Draw a line passing through $D$ and parallel to $AE$. Argue that this line passes through $F$. This helps you show that $AE$ bisects $CD$. Its now not very hard to figure out that $\angle BAC$ is $90$.

 

[anemone]

Re: Find the angle of BAC.

Let the point of trisection of $BC$ nearer to $B$ be $F$. Draw a line passing through $D$ and parallel to $AE$. Argue that this line passes through $F$. This helps you show that $AE$ bisects $CD$. Its now not very hard to figure out that $\angle BAC$ is $90$.

I see it now...the trick is to draw another line parallel to AE that starts from the point D to touch the line of BC.

I will show my full solution here as a way to express my appreciation for your help and guidance...

Join DM so that AE is parallel to DM.
View attachment 631

It follows that $\displaystyle \frac{BD}{DA}=\frac{BM}{ME}$, i.e.

$\displaystyle 1=\frac{BM}{ME}$

$\displaystyle BM=ME$ or $\displaystyle BM=ME=EC $.

It shows that M is the point of trisection of BC nearer to B.

Similarly, by applying the proportionality theorem again, we get

$\displaystyle \frac{CN}{ND}=\frac{CE}{EM}$

$\displaystyle \frac{CN}{ND}=1$

$\displaystyle CN=ND$

This gives us also the fact that $\displaystyle CN=AN$, and $\displaystyle \angle NAC=\angle NCA=\theta $

Last, by adding up all the angles of the triangle ADC and set them equal to $\displaystyle 180^\circ$, we obtain

$\displaystyle 2\theta +2\alpha=180^\circ$

$\displaystyle \theta +\alpha=90^\circ$

$\displaystyle \angle BAC=90^\circ$

Thanks a bunch, caffeinemachine!(Smile)

 

[caffeinemachine]

Re: Find the angle of BAC.

I see it now...the trick is to draw another line parallel to AE that starts from the point D to touch the line of BC.

I will show my full solution here as a way to express my appreciation for your help and guidance...

Join DM so that AE is parallel to DM.
https://www.physicsforums.com/attachments/631

It follows that $\displaystyle \frac{BD}{DA}=\frac{BM}{ME}$, i.e.

$\displaystyle 1=\frac{BM}{ME}$

$\displaystyle BM=ME$ or $\displaystyle BM=ME=EC $.

It shows that M is the point of trisection of BC nearer to B.

Similarly, by applying the proportionality theorem again, we get

$\displaystyle \frac{CN}{ND}=\frac{CE}{EM}$

$\displaystyle \frac{CN}{ND}=1$

$\displaystyle CN=ND$

This gives us also the fact that $\displaystyle CN=AN$, and $\displaystyle \angle NAC=\angle NCA=\theta $

Last, by adding up all the angles of the triangle ADC and set them equal to $\displaystyle 180^\circ$, we obtain

$\displaystyle 2\theta +2\alpha=180^\circ$

$\displaystyle \theta +\alpha=90^\circ$

$\displaystyle \angle BAC=90^\circ$

Thanks a bunch, caffeinemachine!(Smile)

That's great. (Yes)

 

[CellCoree]

Hi there,

It seems like you have made a good attempt at solving this problem. One possible approach to solving this problem is to use the fact that angle ADC is equal to angle BAE. This means that triangles ADC and BAE are similar. From this, we can use the properties of similar triangles to find the measure of angle BAC.

One way to do this is to use the fact that the ratio of corresponding sides in similar triangles is equal. In this case, we can set up the following proportion:

$\displaystyle\frac{AD}{AE} = \frac{CD}{BE}$

Since D is the midpoint of AB, we know that AD is equal to half of AB. Similarly, since E is the point of trisection of BC, we know that BE is equal to two-thirds of BC. Replacing these values in the proportion, we get:

$\displaystyle\frac{\frac{1}{2}AB}{AE} = \frac{CD}{\frac{2}{3}BC}$

Simplifying this, we get:

$\displaystyle\frac{AB}{AE} = \frac{3CD}{2BC}$

Now, we can use the fact that the sum of the angles in a triangle is equal to 180 degrees to set up another proportion:

$\displaystyle\frac{\angle BAC}{\angle CAE} = \frac{AB}{AE}$

Substituting in the values we found in the previous proportion, we get:

$\displaystyle\frac{\angle BAC}{\angle CAE} = \frac{\frac{3CD}{2BC}}{AE}$

We know that angle CAE is equal to 2/3 of angle BAE (since E is the point of trisection of BC). Substituting this and the fact that angle BAE is equal to angle ADC, we get:

$\displaystyle\frac{\angle BAC}{\frac{2}{3}\angle ADC} = \frac{\frac{3CD}{2BC}}{AE}$

Simplifying this, we get:

$\displaystyle\frac{3\angle BAC}{2\angle ADC} = \frac{3CD}{2BC}$

Finally, we can use the fact that angle ADC is equal to angle BAE to simplify this further:

$\displaystyle\frac{3\angle BAC}{2\angle