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Given that $\triangle ABC$ is an isosceles right triangle with $AC=BC$ and angle $ACB=90°$. $D$ is a point on $AC$ and $E$ is on the extension of $BD$ such that $AE$ is perpendicular to $BE$. If $AE=\frac{1}{2}BD$, prove that BD bisects angle $\angle ABC$.

I have tried proving triangle $\triangle AEB$ and triangle $\triangle DCB$ similar but can't do so. After some angle chasing, I arrived at the result that somehow if I prove angle $\angle CDB$ to be $67.5°$ then it could be proved. But I failed to do so.enter image description here

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  • $\begingroup$ Start by drawing a diagram and showing all the given information. $\endgroup$ – 1123581321 Apr 20 at 4:06
  • $\begingroup$ @PushpaKumari just provide a link to your image, someone will be willing to edit it. $\endgroup$ – Quang Hoang Apr 20 at 4:20
  • $\begingroup$ To prove that the two triangles are similar, show that they both have right angles (this is given) and use the fact that angles EDA and BDC are vertically opposite. $\endgroup$ – 1123581321 Apr 20 at 4:24
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A simple geometric solution:

Extend BC and AE to intersect at F. Triangles AFC and BDC are similar. The side CB of triangle BDC is equal to side AC of triangle AFC, this results in that other sides of AFC and BDC are equal including AF and BD and we have $AE=\frac {1}{2}DB=\frac {1}{2}AF$. But AE is also perpendicular to BE, that means BE is the height of ABE and triangle ABF is isosceles and its height BE bisects the angle$ <ABC$.

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  • $\begingroup$ +1. Thinking out of box. $\endgroup$ – farruhota Apr 20 at 6:20
  • $\begingroup$ Mind blown 😱😱 $\endgroup$ – Pushpa Kumari Apr 20 at 8:10
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Refer to the figure:

$\hspace{2cm}$enter image description here

From similarity of triangles $\Delta ADE$ and $\Delta BCD$ (corresponding angles are equal): $$\frac{x}{y}=\frac{y-z}{2x} \Rightarrow 2x^2=y^2-zy \ \ (1)$$ From the right $\Delta BCD$: $$z^2+y^2=(2x)^2 \ \ (2)$$ Now substitute $(1)$ to $(2)$: $$z^2+y^2=2(y^2-zy) \Rightarrow \\ (y-z)^2=2z^2 \Rightarrow \\ y-z=z\sqrt{2} \Rightarrow \\ \frac{y-z}{z}=\frac{y\sqrt{2}}{y},$$ which is consistent with the angle bisector theorem.

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enter image description here

Let $M$ be the midpoint of $BD$ and $G$, $F$ be projections of $M$, $E$ onto $AC$ respectively.

First notice that $\angle GMD = \angle DAE = 90^\circ -\angle ADE$.

Now $\triangle MGD$ and $\triangle AFE$ are congruent as they are right triangle having equal hypothenuse ($MD=AE$) and a pair of equal angles. So $$FA = MG = BC/2 = AC/2.$$ Thus $F$ is the midpoint of $AC$ and $AE = EC$. Since $E$ lies on the circumcircle of $\triangle ABC$, it follows that $E$ is the midpoint of arc $AC$. Hence, $BE$ is the angle bisector of $\angle ABC$.

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