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In an acute-angled triangle ABC with height CD, K and L are orthogonal projections through D respectively on AC and BC. Prove that points A, B, K and L lie on a circle $c$.

I tried to prove that triangles ADK and DLB are of the same form, but without any luck.

KReiser
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  • Hint: the circle will have diameter $AB$. – Wojowu Nov 03 '15 at 20:41
  • I'm not sure what you mean. If you are claiming that $KA=KB$ or $KA=LA$, then this isn't correct. – Wojowu Nov 03 '15 at 20:49
  • @Wojowu The only hit I've got is "All points of which the distance to two given points is a fixed ratio, lie on a circle." –  Nov 03 '15 at 20:51
  • Can you use the fact that the circumcenter of a right triangle is the midpoint of the hypotenuse? If so, consider circumcircles of $ABK$ and $ABL$ and prove these are equal. – Wojowu Nov 03 '15 at 20:53
  • @Wojowu I am not sure what you mean. How do you come to the conclusion that ABK and ABL are right triangles? –  Nov 03 '15 at 20:58
  • Wojowu, I think you misread the problem. $K$ and $L$ are projections of $D$ on $AC$ and $BC$, not the projections of the vertices. – timon92 Nov 03 '15 at 21:00
  • Ah, I'm really sorry, I have misread the problem. In this case I don't think it's necessarily true that $AB$ will be the diameter. – Wojowu Nov 03 '15 at 21:00
  • You can show that $CK \cdot CA = CD^2 = CL \cdot CB$. This implies that $\frac{CA}{CB} = \frac{CL}{CK}$ thus $\triangle CBA \sim \triangle CKL$ – timon92 Nov 03 '15 at 21:08
  • Suppose that I have proved the equal forms between CBA and CKL, I still don't see how this helps me to prove that A, B, K and L lie on a circle c. I don't find the hit I got from the book helpful at all (see one of my previous posts). I would appreciate a solution upon which I could see where do I need to work towards. –  Nov 03 '15 at 21:13
  • Actually they're not congruent but similar. From the similarity you get $\angle CBA = \angle CKL$. – timon92 Nov 03 '15 at 21:16

3 Answers3

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Hint: try to prove the following equalities and conclude from there:

$$\angle KAB=\angle CDK=\angle CLK$$

Wojowu
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First notice that the $\angle CDK=\angle CBA$ because $\angle BKD$ and $\angle CDB$ are both $\pi/2$. For the same reason $\angle CDL=\angle CAB$. From this follows that $$\angle LDK=\angle CDK+\angle CDL=\angle CBA+\angle CAB$$ therefore $$\angle LDK+\angle ACB=\angle CBA+\angle CAB+\angle ACB=\pi.$$ This implies that the points $C,L,D,K$ lie in the same circle. Hence $$\angle DKL= \angle DCL.$$ Now notice that $$\angle LKB+\angle CAB=\angle DKL+\pi/2+\angle CAB=\angle DCL+\pi/2+\angle CAB$$ and from the fact that $\angle DCB+\angle DBC=\pi/2$ we get $$\angle LKB+\angle CAB=\angle DCL+\angle DCB+\angle DBC+\angle CAB=\pi.$$ So $\angle LKB+\angle CAB=\pi$ and $A,L,K,B$ lie in the same circle.

mrprottolo
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$\begin{array}{} B⟂AB & BE∩KD={E} \\ AF⟂AB & AF∩DL={F} \\ \end{array}$

$ΔCDL∼ΔADF$

$\begin{array}{} \frac{CD}{DL}=\frac{DF}{AF} & ⇒ & AF=\frac{DL·DF}{CD} \end{array}$

$ΔCAD∼ΔBDE$

$\begin{array}{} \frac{CD}{AD}=\frac{BD}{BE} & ⇒ & BE=\frac{AD·BD}{CD} \end{array}$

$ΔADF∼ΔBDL$

$\begin{array}{} \frac{AD}{DF}=\frac{DL}{BD} & ⇒ & AD·BD=DL·DF \end{array}$

$\begin{array}{} AF=BE & ⇒ AE=BF & (ΔABE≅ΔABF) \end{array}$

(angle in a semicircle): right triangles:

$\begin{array}{} ΔAKE, ΔABE & AE \text{(diameter)} & \overset{\Huge\frown}{AKBE} \\ ΔABF, ΔBLF & BF\text{(diameter)} & \overset{\Huge\frown}{BLAF} \end{array}$

$\begin{array}{} AE∩BF=M & \text{(Parallelogram Diagonals Theorem)} \end{array}$

semicircles ($\overset{\LARGE\frown}{AKE}$ and $\overset{\LARGE\frown}{BLF})$ with congruent diameters ($AE = BF$) and common centers (midpoint $M$) are arcs contained in circle $M$ (circle $AKLB$)

circleKLB

Wilson Massaro
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