# Find the measure of angle BAC.

#### anemone

##### MHB POTW Director
Staff member
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. 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.

#### caffeinemachine

##### Well-known member
MHB Math Scholar
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.

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.

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

##### MHB POTW Director
Staff member
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. 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! #### caffeinemachine

##### Well-known member
MHB Math Scholar
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.
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! That's great. 