In triangle $ABC$ we have $\angle ACB = 90^o$. The point $M$ is the midpoint of $AB$. The line through $M$ parallel to $BC$ intersects $AC$ in $D$. The midpoint of line segment $CD$ is $E$. The lines $BD$ and $CM$ are perpendicular. (a) Prove that triangles $CME$ and $ABD$ are similar. (b) Prove that $EM$ and $AB$ are perpendicular. [asy][asy] unitsize(1 cm); pair A, B, C, D, E, M; A = (0,0); B = (4,0); C = (2.6,2); M = (A + B)/2; D = (A + C)/2; E = (C + D)/2; draw(A--B--C--cycle); draw(C--M--D--B); dot("$A$", A, SW); dot("$B$", B, SE); dot("$C$", C, N); dot("$D$", D, NW); dot("$E$", E, NW); dot("$M$", M, S); [/asy][/asy] Be aware: the figure is not drawn to scale.
Problem
Source: Dutch NMO 2021 p4
Tags: geometry, similar triangles, perpendicular, right triangle
Albert123
02.01.2022 09:52
(a) The triangles $CME$ and $ABD$ are similar with factor $\frac{1}{2}$.$\blacksquare$ (b) Let $\angle CAM=\angle ACM=\alpha \implies \angle ABD=90-2\alpha \implies \angle MDB=\alpha \implies \angle EDB=\angle DEM=90-\alpha \implies EM \perp AB$.$\blacksquare$
Theavenger1945
19.02.2022 21:34
Tsikaloudakis (Profile) $\cdot$ 11 minutes ago $\cdot$ Y 0 $\cdot$
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βλέπε σχήμα:
βλέπε σχήμα: means see figure btw
Noob_at_math_69_level
21.02.2022 06:13
Let $CM$ meet $BD$ at $F$. Let $B'$ be the reflection of $B$ by $E$. $a)$ We can see that $B',D,M$ collinear. (Because $\angle B'DE + \angle CDM = \angle BCE + \angle DCB= 180º$) Since $B'DBC$ is a parallelogram ($E$ mid point $DC$,$BB'$) $=>$ $B'C=DB$. Since $D$ is the mid point of $AC$ and $\angle ADM = 90º$ $=>$ $A$ is the reflection of $C$ by $D$. $=>$ $B'A=B'C=DM$. So $B'A$ = $2EM$ = $DB$ $(1)$ $\angle ACB = 90º$, $M$ mid point $AB$ $=>$ $2CM$ = $AB$. $(2)$ $2CE$ = $CD$ = $AD$ $(3)$. From $(1),(2),(3)$ $=>$ Triangle $CME,ABD$ similar $(s.s.s)$ $b)$ Since $DB//B'C$ ,$BD$ and $CM$ perpendicular $=>$ $\angle B'CM = 90º=\angle B'AM = \angle EMB$ $=>$ $EM$ and $AB$ perpendicular
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