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Answer by Lion Heart for Find the segment "DC" in the obtuse triangle below
Let $\angle BAD = \angle CAD = \alpha$ then $\angle BAD =\angle BCA = 2\alpha$Draw $DE$ such that $AE=QE$ ($E$ is the midpoint of $AQ$)then $AE=QE=DE=5$ and $\angle ADE = \angle DAE = \alpha$then...
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Let $R$ be the midpoint of $AQ$, so $\lvert AR\rvert = \lvert RQ\rvert = 5$. Since $\triangle ADQ$ is right-angled, then $AQ$ is the diameter of its circumcircle, so $R$ is its center, which means that...
View ArticleAnswer by John Omielan for Find the segment "DC" in the obtuse triangle below
Using your result of $\measuredangle CQD = 90^{\circ} + \theta$ and the Law of sines with $\triangle CDQ$ gives$$\begin{equation}\begin{aligned}\frac{\lvert DC\rvert}{\sin(90^{\circ} + \theta)} & =...
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Find the segment "DC" in the obtuse triangle below
In an obtuse triangle $ABC$, obtuse at $B$, the internal bisector $AD$ is drawn and in $AC$ the point $"q"$ is taken such that $m∡𝐴𝐷𝑄=90^𝑜$.Calculate $DC$. If: $AQ = 10$ and $AB = BC$.$Answer:10$I...
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