$ABC$ is a right angle triangle where $A$ is the right angle. $D$ is a point on $AC$ so
that $AB=AD$. $E$ & $F$ bisect $BD$ & $AD$ respectively. $DH \bot BC$ & $DH=\sqrt{2}$. Find the $\angle DCH$ when $EF=1$.
Dhaka Secondary 2009/3
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Please don't post problems (by starting a topic) in the "Secondary: Solved" forum. This forum is only for showcasing the problems for the convenience of the users. You can post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
Please don't post problems (by starting a topic) in the "Secondary: Solved" forum. This forum is only for showcasing the problems for the convenience of the users. You can post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
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Tahmid Hasan
- Posts:665
- Joined:Thu Dec 09, 2010 5:34 pm
- Location:Khulna,Bangladesh.
Re: Dhaka Secondary 2009/3
DHC and ABC are similar.\[so \frac{CD}{BC}=\frac{CH}{AC}=\frac{DH}{AB}=\frac{1}{\sqrt{2}}\]
from this ,we need to make \[\frac{CD}{CH} or \frac{AB}{AC}\]
or anything else and may find C with trigonometric ratios.
it's my way(may be same) but how to value these factions?that makes to do 1st similarity eqtn ''ekantorkoron''
it then changes value.please,help.
from this ,we need to make \[\frac{CD}{CH} or \frac{AB}{AC}\]
or anything else and may find C with trigonometric ratios.
it's my way(may be same) but how to value these factions?that makes to do 1st similarity eqtn ''ekantorkoron''
it then changes value.please,help.
Try not to become a man of success but rather to become a man of value.-Albert Einstein