Isosceles Triangle And Peculiar Point Reloaded

Geometry Level 4

$ABC$ is an isosceles triangle with $AB=AC$ , $\angle BAC = 96 ^\circ$ . $D$ is a point such that $\angle ACD = 48 ^\circ$ , $AD = BC$ and angle $DAC$ is obtuse. What is the measure (in degrees) of $\angle DAC$ ?

Inspiration .

The answer is 102.

2 solutions

Ahmad Saad
Mar 31, 2016

Consider a point $E$ such that $AE\perp BC$ , then $E$ is a midpoint of $BC$ .

Similarly, consider a point $F$ such that $AF\perp CD$ , then $\angle CAF = 90^\circ - 48^\circ = 42^\circ$ .

Figure above shows that the two triangles $AFC$ and $BEA$ are similar.

Then, $\dfrac{AF}{BE} = \dfrac{CA}{AB} = 1 \quad (\because AB=AC)$ .

Therefore, $AF = BE = \dfrac12 BC = \dfrac12 AD \quad (\because AD=BC)$ .

Triangle $AFD$ is right angle at $F$ , thus $\angle DAF = 60^\circ$ .

$\angle DAC = 60^\circ + 42^\circ = 102^\circ$ .

Just to point out, what is the difference between this problem and the problem from which the inspiration was?

Shourya Pandey - 5 years, 2 months ago

The original problem:

the original

The inspired problem:

the insprired

Abdelhamid Saadi - 5 years, 2 months ago

Angle DAC is obtuse in both cases. If there is no difference in the wording of the problem, how can the diagrams be different?

Shourya Pandey - 5 years, 2 months ago

@Shourya Pandey – There is a slight difference in the wording of the problem.

In the original says : $\angle DCA = 48 ^\circ$

In the inspired says : $\angle ACD = 48 ^\circ$

While it's the same angle, it could be oriented either clockwise or anticlockwise.

Abdelhamid Saadi - 5 years, 2 months ago

@Abdelhamid Saadi – But you could draw ABC clockwise or anti-clockwise, so I don't get what convention you chose.

Shourya Pandey - 5 years, 2 months ago

@Shourya Pandey – I think there is no universal convention for oriented angles.

I follow the same convention as in GeoGebra.

Abdelhamid Saadi - 5 years, 2 months ago

Although you claim that the problems are different, I solved both in exactly the same way.

A Former Brilliant Member - 5 years, 2 months ago

Your proof works for both cases, and with the same result. By this problem I want to point out this symmetry.

But not all the proofs will work for both cases.

A very clever proof by Calvin Lin won't work directly without a little tweak.

Abdelhamid Saadi - 5 years, 2 months ago

Khoa Đăng
May 10, 2016

Isosceles Triangle And Peculiar Point Reloaded

The answer is 102.

2 solutions

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