What Is This Line?

Geometry Level 4

In $\triangle ABC,$ $AB= 6, BC= 7, CA= 5.$ There exists a unique line $\ell$ passing through $A$ such that reflections of $B,C$ about $\ell$ lie on lines $CA, AB$ respectively. Suppose $\ell$ intersects $BC$ at point $D.$ If $\dfrac{BD}{DC}= \dfrac{a}{b}$ for some coprime positive integers $a, b,$ find $a+b.$

Details and assumptions
- The reflections of $B,C$ in $\ell$ lie on the lines $CA, AB$ respectively. They might not lie on the segments $CA, AB.$
- In the diagram above, $B', C'$ are the reflections of $B,C$ respectively.
- The diagram shown is not accurate.

The answer is 11.

1 solution

Sreejato Bhattacharya
Apr 17, 2014

Let $BB'$ meet $\ell$ at $X.$ By virtue of reflection, we have $BX= XB', \angle BXA = \angle B'XA,$ so $\triangle ABX \cong \triangle AB'X.$ This gives $\angle BAD = \angle DAC,$ so $AD$ is the internal angle bisector of $\angle A.$ In a similar manner, we can also use the fact that $C'$ lies on $AB$ and get the same conclusion. The exact same arguments prove the converse statement: if $AD$ bisects $\angle BAC,$ $B',C'$ lie on $CA, AB$ respectively. We have $\dfrac{BD}{DC}= \dfrac{AB}{AC}= \dfrac{6}{5},$ so $a+b= \boxed{11}.$

A picture will accompany soon.

$\dfrac{BD}{DC}=\dfrac{AB}{AC}=\dfrac{6}{5}$ , but your point is made.

Daniel Liu - 7 years, 1 month ago

Thanks for pointing it out!

Sreejato Bhattacharya - 7 years, 1 month ago

(BD/DC)=(a/b); so ((BD+DC)/DC)=((a+b)/b) there BC=7 so a+b=7........but my answer doesn't match with the answer above.please help me by defining the fault of my thinking

Shahriar Manna - 7 years, 1 month ago

The question doesn't ask for $\dfrac{BD+DC}{DC}.$

Sreejato Bhattacharya - 7 years, 1 month ago

What Is This Line?

The answer is 11.

1 solution

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