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The base angles of a triangle are $22.5^\circ$ and $112.5^\circ$. Show that the base is twice the height.

My Attempt enter image description here

$$ h=c.\sin22.5^\circ=c.\cos 67.5^\circ\\ =b\sin 67.5^\circ=b\cos 22.5^\circ $$ $$ a=c\cos22.5^\circ- b\sin22.5\circ=\frac{h}{b}-\frac{h}{c}=h\cdot\frac{c-b}{bc} $$

I have no clue of how to prove this.

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  • $\begingroup$ You should have pointed upon every vertex. $\endgroup$ – Rakibul Islam Prince Jan 1 at 23:15
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    $\begingroup$ @RakibulIslamPrince Fixed. $\endgroup$ – A.Γ. Jan 1 at 23:51
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By the law of sines, $$\frac{a}{\sin{45^{\circ}}}=\frac{b}{\sin{22.5^{\circ}}}$$

By the double angle formula, this is equivalent to $$\frac{a}{2\sin{22.5^{\circ}}\cos{22.5^{\circ}}}=\frac{b}{\sin{22.5^{\circ}}}\implies\frac{a}{2\cos{22.5^{\circ}}}=b$$

From the smaller right triangle we see that $$\frac{h}{b}=\cos{22.5^{\circ}}\implies h=b\cos{22.5^{\circ}}$$

Combining the results gives $a=2h$.

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The two right angled triangles in your picture are similar and both have smaller angle $22.5$. Let the shortest unmarked side be $x$

Then $$\frac{h}{a+x}=\frac xh=\tan22.5=\sqrt{2}-1$$

Eliminating $x$ gives $$h^2=ah(\sqrt{2}-1)+h^2(\sqrt{2}-1)^2$$ Rearranging gives $$\frac ah=\frac{2\sqrt{2}-2}{\sqrt{2}-1}=2$$

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Let,the extended portion of $a=x$ and let $D$ be the intersection point of base and height.

So,$CD=x$

Now,in the $\triangle ABD$, $$\tan (22.5)=\frac{h}{a+x}\implies h=a\tan(22.5)+x\tan(22.5)......(1)$$ in the $\triangle ACD$, $$\tan (67.5)=\frac{h}{x}\implies h=x\tan (67.5)\implies x=\frac{h}{\tan (67.5)}$$ from (1), $$h=a\tan(22.5)+\frac{h}{\tan (67.5)}\tan(22.5)$$ $$\implies h=\frac{a}{2}[\text{after simplification}]$$

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The base of the triangle is $c\cos(22.5)-b\cos(180-112.5)$. The height is $c\sin(22.5)$. Also, $b\sin(180-112.5) = c\sin(22.5)$. Use these two relations to write $b$ in terms of $c$ and show that $$\frac{c\sin(22.5)}{c\cos(22.5)-b\cos(180-112.5)} = 1/2.$$ (Hint: the $c$ will cancel out).

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Solution without trigonometry.

Since $$\measuredangle CAD=\measuredangle BCD-90^{\circ}=112.5^{\circ}-90^{\circ}=22.5^{\circ}=\measuredangle ABD,$$ we get that $DA$ is a tangent line to the circumcircle of $\Delta ABC$.

Let $O$ be a center of the circle and $OM$ be an altitude of $\Delta OBC$.

Since $OA\perp DA$, we obtain $$\measuredangle $OCM=\measuredangle ABC-\measuredangle OCA=\measuredangle ABC-\measuredangle OAC=112.5^{\circ}-(90^{\circ}-22.5^{\circ})=45^{\circ},$$ which says $OM=MC$ and since $BM=MC,$ we obtain: $$BC=2OM=2AD$$ and we are done!

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