What´s measure of the $\angle MNQ$ in the figure below euclidean-geometryplane-geometry For reference: Calculate the measure of the $\angle MNQ$ If: H ➔ Orthocenter M ➔ Midpoint of AB N ➔ Midpoint of BC Q ➔ Midpoint of AH Follow the drawing with the relationships I found Aqua Solution Image: Best Answer Hint: Points $M,N,K,Q$ are concyclic. Why? Nine point circle So, we have: $$\angle MNQ = \angle MKQ = \angle MAK = 90^{\circ}- \angle B= 10^{\circ}$$ Related SolutionsWhat´s the measure of angle $\angle IMN$ Let $BH$ intersect $MI$ and $MN$ at $P$ and $Q$ respectively. In two similar triangles, ratio of corresponding sides=ratio of inradii $\small{\triangle ABH\sim\triangle BCH}\implies\frac{AB}{BC}=\frac{GH}{HF}=\frac{MH}{HN}\implies\small{\triangle ABC\sim\triangle MHN}$ Therefore, $\angle HMN=90^\circ-\theta$ $\angle MQP=45^\circ+(90^\circ-\theta)=135^\circ-\theta$ $\angle ABH=\theta, \angle BAD=45^\circ-\frac{\theta}{2}\implies\angle BPM=135^\circ-\frac{\theta}{2}$ Therefore from $\triangle PQM$, $\angle IMN=(135^\circ-\frac{\theta}{2})-(135^\circ-\theta)=\frac{\theta}{2}$ What’s the measure of the $\angle ACB$ From $I$ and $E$ drop perpendiculars to $AC$. These perpendiculars will be inradius $r$ and exradius, $r_a$ respectively. Since $AI=AE/2$, $r=r_a/2$ using similar triangles. Using formulas for $r=\Delta / s$ and $r_a = \Delta / (s-a)$, we can obtain $$b+c=3a$$ Combining this with $b^2-c^2=a^2$, we get $$\frac{c}{b}=\frac{4}{5} \Rightarrow \angle C = \sin^{-1}\frac{4}{5} \approx 53^\circ$$ Related QuestionIn the triangle ABC, what’s the measure $\angle ACB$?.What’s the measure of the $\angle BAC$ in the question belowWhat’s the measure of angle $PCB$ in the figure below
Best Answer
Hint: Points $M,N,K,Q$ are concyclic. Why?
So, we have: $$\angle MNQ = \angle MKQ = \angle MAK = 90^{\circ}- \angle B= 10^{\circ}$$