In right triangle $ABC(\angle C=90)$, points $K,L,M$ lie on $AC,BC,AB$ respectively,prove that: $AK=KM$

Question

In right triangle $ABC(\angle C=90)$, points $K,L,M$ lie on $AC,BC,AB$ respectively, so that: $AK=BL,KM=LM,\angle KML=90$.Prove that: $AK=KM$

We know $C,K,M,L$ lie on the same circle. Triangles $AKM$ and $BLM$ don't seem to be identical! We should show that $\angle KAM=\angle KMA$...

Answer 1

Let's take a look at $\triangle AKM$, $\triangle BML$ and the cyclic quadrilateral $CKML$. By the Sine Law, $$\frac{KM}{\sin A} = \frac{AK}{\sin\angle KMA} \implies \frac{KM}{AK} = \frac{\sin A}{\sin\angle KMA}$$ and $$\frac{BL}{\sin \angle LMB} = \frac{ML}{\sin B} \implies \frac{ML}{BL} = \frac{\sin B }{\sin \angle LMB} = \frac{\cos A}{\cos \angle KMA}$$ Because $\angle A + \angle B = 90^\circ$, $\angle LMB + \angle KMA = 90^\circ$. Hence, $$\frac{\sin A}{\sin\angle KMA} = \frac{\cos A}{\cos \angle KMA} \implies \tan A = \tan \angle KMA \implies KM = KA$$ as desired.

Note: the last step follows because if we draw a perpendicular line $KT\perp AM$ at $T$, then $$\tan A = \tan KMA \implies \frac{KT}{TM} = \frac{KT}{AT}\implies MT = AT$$ which with $KT\perp AM$ gives us $\triangle KMT \cong \triangle KAT\implies KM = KA$.

Answer 2

The points $C, K, M, L$ are concyclic, hence $\angle LCM = \angle LKM = 45^{\circ}$. Thus $M$ is on the bisector of $\angle ACB$. It follows that $M$ is equidistant from sides $CA$ and $CB$.

Therefore, if we draw any two perpendicular lines emanating from $M$ (not parallel to the legs of $ABC$), and we let them intersect $CA$ and $CB$ at $K'$ and $L'$, then we have $K'M = L'M$. Moreover, the sum $AK' + BL'$ is independent of the choice of perpendicular lines.

We pick two such lines so that $\angle K'MA = \angle CAB$ and $\angle L'MB = \angle CBA$. Then triangles $K'MA$ and $L'MB$ are isosceles, so we have $AK' = K'M = L'M = BL'$.

The points $K$ and $L$ are obtained by selecting a different pair of perpendicular lines through $M$. But since $AK + BL = AK' + BL'$, $AK = BL$ and $AK' = BL'$, we must have $K = K'$ and $L = L'$.

Therefore $AK = KM$.

Answer 3

Let $P\in BC$ such that $L$ is a midpoint of $BP$.

Since $\measuredangle PLM=\measuredangle MKA$, we see that $\Delta PLM\cong\Delta AKM$.

Thus, $\measuredangle P=\measuredangle A$ and from here $\measuredangle P+\measuredangle B=90^{\circ}$ or $\measuredangle PMB=90^{\circ}$.

But $ML$ is a median of $\Delta PMB$, which says $PL=BL=ML$ and we are done!