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Let a circle (O) and a point A outside of the circle. AB, AC are tangents of (O) (B,C $\in$ (O)). BD is an diameter of (O). CK perpendicular to BD (K $\in$ BD). Let I is intersection of CK and AD. Prove that I is midpoint of KCenter image description here

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2  
midpoint of what? $[K,C]$ I guess!? –  Mercy Dec 11 '12 at 15:56
    
yes:) can you help me? –  Le Van Tu Dec 11 '12 at 17:09
    
What have you tried so far? –  chubbycantorset Dec 11 '12 at 17:31

3 Answers 3

enter image description here

Without any loss of generality we can assume $B(a,0),O(0,0)$ so that $D(-a,0)$ the equation of the circle $(x-0)^2+(y-0)^2=(a-0)^2\implies x^2+y^2=a^2$

Using this, the equation of the tangent at any point $P(a\cos\theta,a\sin\theta)$ will be $xa\cos \theta+ya\sin\theta=a^2\implies x\cos \theta+y\sin \theta=a$

So, $AB: x=a--->(1)$ (putting $\theta=0$)

Let $C$ be $(a\cos \beta,a\sin \beta)$ so $CA: x\cos\beta+y\sin\beta=a--->(2)$ (putting $\theta=\beta$)

Solving for $y$ we get, $y=\frac{a(1-\cos\beta)}{\sin \beta}$ so $A(a,\frac{a(1-\cos\beta)}{\sin \beta})$

$AD$ will be $$\frac{y-0}{x+a}=\frac{0-\frac{a(1-\cos\beta)}{\sin \beta})}{-a-a}=\frac{1-\cos\beta}{2\sin \beta}\implies 2y=(x+a)\frac{1-\cos\beta}{\sin \beta}$$

Clearly, the abscissa of $I$ will be $a\cos \beta$

So, $2y=(a\cos \beta+a)\frac{1-\cos\beta}{\sin \beta}=a\sin\beta\implies y=\frac{a\sin\beta}2$

Hence, $I(a\cos\beta,\frac{a\sin\beta}2)$ is the midpoint of $C(a\cos\beta,a\sin\beta)$ and $K(a\cos\beta,0)$

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This might help: Notice that $OKC$ is a 30-60-90 triangle, and so is $ABJ$, where $J$ is the midpoint of $BC$, as $ABC$ is equilateral. Also note that angle $ICA$ is 120 degrees (check it!). Now, draw a new line segment from $C$ to $AB$ with the following property: the point $L$, where the new line segment intersects $AD$ should be such that angle $DLC$ = angle $AIC$, i.e. triangle $LIC$ is isoceles. When you've created such a point, note what it does to line segment $AJ$ (Does it bisect it?). What can you tell about what point $I$ does to segment $KC$?

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There are two way to prove this problem.

Prove 1: enter image description here

Because BD is diameter of (O) so CD is perpendicular to BC, then DK=$\dfrac{CD^2}{BD}$

In the other hand, CK//AB so $\dfrac{HK}{AB}=\dfrac{DK}{BD}=\dfrac{CD^2}{BD^2} \Rightarrow HK=\dfrac{CD^2}{BD^2}.AB \, (1)$\ CD is perpendicular to BC so CK.BD=BC.CD $\Rightarrow CK=\dfrac{CD}{BD}.BC \, (2)$

Because $\Delta BCD \sim \Delta ABO \Rightarrow \dfrac{BC}{AB}=\dfrac{CD}{OB}=2.\dfrac{CD}{BD} \Rightarrow BC=2.\dfrac{CD}{BD}.AB \, (3)$

From (2),(3):$CK=\dfrac{CD}{BD}.2.\dfrac{CD}{BD}.AB=\dfrac{CD^2}{BD^2}.AB=2HK$

Prove 2:enter image description here

Let DH is tangent of (O) (H $\in$ AC). J is intersection of BH and AD, ww'll prove that CJ is perpendicular to BD so I $ \equiv$ J.

DH and AB are tangent of (O), BD is a diameter so $AB // DH$, J is intersection of AD and BH so $\dfrac{DJ}{AJ}=\dfrac{DH}{AB}$,

We also have AB=AC, HD=HC so $\dfrac{DJ}{AJ}=\dfrac{HC}{CA} \Rightarrow CJ // DH$ so CJ is perpendicular to BD. So we have I $\equiv$ J ( is both intersection of the parallel to AB across C and AD).

Now we'll prove that I is mid point of CK. We have $CK//AB, CK//DH$ so $\dfrac{CI}{DH}=\dfrac{AI}{ID}=\dfrac{DI}{BI}=\dfrac{IK}{DH} \Rightarrow CI=KI$ (q.e.d)

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