Full question: Let $\triangle$ ABC be a isosceles triangle (AB = AC) and its area is 501 $cm^2$. BD is the internal bisector of $\widehat{ABC}$ (D $\in$ AC). E is a point on the opposite ray of CA such that CE = CB. I is a point on BC such that CI = 1/2 BI. EI intersects AB at K. BD intersects KC at H. Find $Area_{ \triangle AHC}$.
The figure should look like this:
Answer: 167 $cm^2$
I found this question on a national competition, which ended in May 2016
My effort: Let HN $\bot$ BC, HM $\bot$ AB, IL $\bot$ AC, AO $\bot$ BC
The figure should be:
Then the right triangles $\triangle$ BMH and $\triangle$ BNH are congruent (Hypotenuse-Angle theorem), so HM = HN.
$$Area_{\triangle AHC} = Area_{\triangle ABC} - Area_{\triangle AHB} - Area_{\triangle BHC}$$
$$Area_{\triangle AHC} = 501 - \frac{1}{2}.HM.AB - \frac{1}{2}.HN.BC$$
$$Area_{\triangle AHC} = 501 - \frac{1}{2}.HN.(AB + BC)\\ = 501 - \frac{1}{2}.HN.(AC + CE)\\ = 501 - \frac{1}{2}.HN.AE$$
Here comes the 'crazy' part. A friend told me she could solve HN.AE = 2.IL.AC, then:
$$Area_{\triangle AHC} = 501 - 2.\left(\frac{1}{2}.IL.AC\right)\\ = 501 - 2.Area_{\triangle AIC}$$
$$CI = \frac{1}{2}.BI \implies \frac{CI}{BC} = \frac{CI}{CI + BI} = \frac{\frac{1}{2}.BI}{\frac{1}{2}.BI + BI} = \frac{1}{3}$$
We then have:
$$\frac{Area_{\triangle AIC}}{Area_{\triangle ABC}} = \frac{\frac{1}{2}.AO.IC}{\frac{1}{2}.AO.BC} = \frac{IC}{BC} = \frac{1}{3} \implies Area_{\triangle AIC} = \frac{1}{3}.Area_{\triangle ABC} = \frac{1}{3}.501 = 167 (cm^2)$$
So $Area_{\triangle AHC} = 501 - 2.167 = 167 (cm^2)$
- 167 $cm^2$ IS the correct answer, but my friend forgot how to solve HN.AE = 2.IL.AC, which I couldn't myself. So can you help me with that, or am I going the wrong direction?
Thank you for checking in. Any help would be appreciated.
The areas of triangles $AHC$ and $ABC$ are in the ratio $HD/BD$, so I think the easiest way to find the required area is that of computing $HD/BD$.
From $K$ draw a line parallel to $AC$, intersecting $BD$ and $BC$ at $G$ and $F$ respectively (see picture below). Triangles $KFI$ and $ECI$ are similar, which entails $FI={1\over3}KF={1\over3}BK$ and $$ BF=BI-FI={2\over3}BC-{1\over3}BK. $$
By the similarity of $ABC$ and $KBF$ we have $BG=(BF/BC)BD$. Inserting here the above equality yields: $$ BG=\left({2\over3}-{1\over3}{BK\over BC}\right)BD. $$
By the similarity of $KGH$ and $CDH$ we also have $$GH={HK\over HC}HD={BK\over BC}HD,$$ where we have used that ${HK/HC}={BK/BC}$ by the theorem of triangle bisector. In the end we thus get: $$ BD-HD=BH=BG+GH=\left({2\over3}-{1\over3}{BK\over BC}\right)BD+ {BK\over BC}HD, $$ whence $HD={1\over3}BD$.