Everything seems to fit

Geometry Level 4

ABC is a triangle with angle C=70 degrees. Angle bisectors of BX and CY are drawn cutting AC and AB at D and E respectively. Given that Y,A,X are collinear points ,YX is parallel to BC and YE= XD, find angle A.

The answer is 40.

1 solution

A Former Brilliant Member
Dec 30, 2019

From the given conditions we see that $\triangle {ABD}$ and $\triangle {ACE}$ are similar. Hence $\angle {AEC}=\angle {ADB}$ . That is, $35\degree +\angle {ABC}=70\degree+\dfrac{1}{2}\angle {ABC}$ or $\angle {ABC}=70\degree$ and hence $\angle {BAC}=180\degree-140\degree=\boxed {40\degree}$

How did you conclude ABD and ACE are similar triangles?

Shubham Raj - 1 year, 5 months ago

$\overline {AB}\times \overline {BD}=\overline {AC}\times \overline {CE}=\overline {XD}\times \overline {BC}$

A Former Brilliant Member - 1 year, 5 months ago

But for the triangles to be similar, you need AB/BD = AC/CE. Correct me if I am wrong

Shubham Raj - 1 year, 5 months ago

$\triangle {}$ ' $^s$ $EBC$ and $EAY$ are similar. So $\dfrac{\overline {BE}}{\overline {AE}}=\dfrac{\overline {BC}}{\overline {AC}}=\dfrac{\overline {CE}}{\overline {YE}}$ . $\triangle {}$ ' $^s$ $DBC$ and $ADX$ are similar. So $\dfrac{\overline {CD}}{\overline {AD}}=\dfrac{\overline {BD}}{\overline {XD}}={\overline {BC}}{\overline {AB}}=\dfrac{\overline {BD}}{\overline {YE}}$ . So, $\dfrac{\overline {AB}}{\overline {AC}}=\dfrac{\overline {CE}}{\overline {BD}}$ or $\overline {AB}\times \overline {BD}=\overline {AC}\times \overline {CE}$ . Or $\dfrac{\overline {AB}}{\overline {AC}}=\dfrac{\overline {CE}}{\overline {BD}}$ . Since $\triangle {ABC}$ is isosceles, $\overline {BD}=\overline {CE}$

A Former Brilliant Member - 1 year, 5 months ago

I concur with the fact that AB BD = AC CE. But again, it is not given thay ABC is isosceles.

Shubham Raj - 1 year, 5 months ago

ABx BD = AC X CE is what I meant

Shubham Raj - 1 year, 5 months ago

Everything seems to fit

The answer is 40.

1 solution

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