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# 8.7: More exercises

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Exercise $$\PageIndex{1}$$

Assume that an angle bisector of a nondegenerate triangle bisects the opposite side. Show that the triangle is isosceles.

Hint

Apply Lemma 8.4.1. Also see the solution of Exercise 11.1.1.

Exercise $$\PageIndex{2}$$

Assume that at one vertex of a nondegenerate triangle the bisector coincides with the altitude. Show that the triangle is isosceles.

Hint

Apply ASA to the two triangles that the bisector cuts from the original triangle.

Exercise $$\PageIndex{3}$$

Assume sides $$[BC]$$, $$[CA]$$, and $$[AB]$$ of $$\triangle ABC$$ are tangent to the incircle at $$X$$, $$Y$$, and $$Z$$ respectively. Show that

$$AY = AZ = \dfrac{1}{2} \cdot (AB + AC - BC).$$ By the definition, the vertexes of orthic triangle are the base points of the altitudes of the given triangle.

Hint

Let $$I$$ be the incenter. By SAS, we get that $$\triangle AIZ \cong \triangle AIY$$. Therefore, $$AY = AZ$$. The same way we get that $$BX = BZ$$ and $$CX = CY$$. Hence the result.

Exercise $$\PageIndex{4}$$

Prove that the orthocenter of an acute triangle coincides with the incenter of its orthic triangle.

What should be an analog of this statement for an obtuse triangle?

Hint

Let $$\triangle ABC$$ be the given acute triangle and $$\triangle A'B'C'$$ be its orthic triangle. Note that $$\triangle AA'C \sim \triangle BB'C$$. Use it to show that $$\triangle A'B'C \sim \triangle ABC$$.

The same way we get that $$\triangle AB'C' \sim \triangle ABC$$. It follows that $$\measuredangle A'B'C = \measuredangle AB'C'$$. Conclude that $$(BB')$$ bisects $$\angle A'B'C'$$.

If $$\triangle ABC$$ is obtuse, then its orthocenter coincides with one of the excenters of $$\triangle ABC$$; that is, the point of intersection of two external and one internal bisectors of $$\triangle ABC$$.

Exercise $$\PageIndex{5}$$

Assume that the bisector at $$A$$ of the triangle $$ABC$$ intersects the side $$[BC]$$ at the point $$D$$; the line thru $$D$$ and parallel to $$(CA)$$ intersects $$(AB)$$ at the point $$E$$; the line thru $$E$$ and parallel to $$(BC)$$ intersects $$(AC)$$ at $$F$$. Show that $$AE = FC$$. Hint

Apply Theorem 4.3.1, Theorem 7.3.1 and Lemma 7.5.1.