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

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

My Attempt

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

I have no clue of how to prove this.

• You should have pointed upon every vertex. – Rakibul Islam Prince Jan 1 at 23:15
• @RakibulIslamPrince Fixed. – A.Γ. Jan 1 at 23:51

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 $$a=2h$$.

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 $$\frac ah=\frac{2\sqrt{2}-2}{\sqrt{2}-1}=2$$

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), $$h=a\tan(22.5)+\frac{h}{\tan (67.5)}\tan(22.5)$$ $$\implies h=\frac{a}{2}[\text{after simplification}]$$

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).

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!