The diagonals of the trapezoid $ABCD$ are perpendicular ($AD//BC, AD>BC$) . Point $M$ is the midpoint of the side of $AB$, the point $N$ is symmetric of the center of the circumscribed circle of the triangle $ABD$ wrt $AD$. Prove that $\angle CMN = 90^o$. (A. Mudgal, India)
Problem
Source: 2018 Oral Moscow Geometry Olympiad grades 10-11 p2
Tags: geometry, trapezoid, perpendicular, diagonals, right angle
MP8148
25.07.2019 06:14
We use complex numbers. WLOG let $|a| = |b| = |d| = 1$. Since $\overline{AD} \parallel \overline{BC}$, we have $$\dfrac{c-b}{d-a} = \overline{\left(\dfrac{c-b}{d-a}\right)} \implies \overline{c} = \dfrac{b-c}{ad} + \dfrac{1}{b}.$$Since $\overline{AC} \perp \overline{BD}$, we have $$\dfrac{c-a}{b-d} + \overline{\left(\dfrac{c-a}{b-d}\right)} = 0 \implies \overline{c} = \dfrac{c-a}{bd} + \dfrac{1}{a}.$$Solving, we have $$\dfrac{b-c}{ad} + \dfrac{1}{b} = \dfrac{c-a}{bd} + \dfrac{1}{a} \implies c = \dfrac{a^2+b^2+d(a-b)}{a+b}.$$We also have $m = \dfrac{a+b}{2}$ and $n = a+d$, so we can verify $$\dfrac{c-m}{n-m} = \dfrac{\tfrac{a^2+b^2+d(a-b)}{a+b} - \tfrac{a+b}{2}}{a+d-\tfrac{a+b}{2}} = \dfrac{2(a^2+b^2+d(a-b)) - (a+b)^2}{2(a+b)(a+d) - (a+b)^2} = \dfrac{(a-b)(a-b+2d)}{(a+b)(a-b+2d)} = -\overline{\left(\dfrac{c-m}{n-m}\right)},$$and we are done.
andrenguyen
25.07.2019 07:58
Let $O$ is the circumcenter of triangle $ABD$. $P$ is the midpoint of segment $AD$. $Q$ is the reflection of $O$ wrt $AB$. It is obvious that $\angle BAC = 90^o - \angle ABD = \angle OAP = \angle OMP = \angle OQN$ Also, $\angle CBA = 180^0 - \angle BAD = \angle NOQ$ Therefore, $\triangle NOQ \sim \triangle CBA$. We have $M$ is the midpoint of segments $AB$ and $OQ$ Then $\triangle CBM \sim \triangle NOM$. Hence $\angle CMB = \angle NMO$ hence $\angle CMN = \angle BMO = 90^o$.
Attachments:
Mahdi_Mashayekhi
29.12.2021 19:31
Let M,S be midpoints of AB,AD. Let K be reflection of O about M. ∠CBA = 180 - ∠A = ∠SOM = ∠NOK and ∠BAC = 90 - ∠B = ∠OAS = ∠OMS = ∠OKN. triangles KON and ABC are similar and M is midpoint of Both KO and AB so NMO and CMB are similar. ∠OMN = ∠BMC ---> ∠CMN = ∠BMO = 90. we're Done.