Question

第一题
已知：△abc中，ab = bc，d是ac的中点，过d作de⊥bc于e，连接ae，取de中点f，连接bf。求证：ae⊥bf
简证：rt△bde∽rt △dce ，
\\(\frac{bd}{de} = \frac{dc}{ce}\\)
\\(\frac{bd}{df} = 2\frac{bd}{de} = 2\frac{dc}{ce} = \frac{ac}{ce}\\)，△bdf∽△ace
∠dbf = ∠cae，∴a、d、g、b四点共圆。
∠bga = ∠bda = 90°，ae⊥bf。

Explanation:

Step1: Analyze triangle properties

In \(\triangle ABC\), \(AB = BC\), \(D\) is the mid - point of \(AC\). By the property of isosceles triangles (the median from the apex of an isosceles triangle is also the altitude and the angle - bisector), we have \(BD\perp AC\), so \(\angle BDC = 90^{\circ}\). Also, since \(DE\perp BC\), \(\angle DEC=\angle DEB = 90^{\circ}\).

In \(Rt\triangle BDE\) and \(Rt\triangle DCE\), \(\angle BDE+\angle CDE = 90^{\circ}\), \(\angle C+\angle CDE=90^{\circ}\), so \(\angle BDE=\angle C\). According to the AA (Angle - Angle) similarity criterion for right - angled triangles, \(Rt\triangle BDE\sim Rt\triangle DCE\). Then we can get the proportion \(\frac{BD}{DE}=\frac{DC}{CE}\).

Step2: Use the mid - point property

Since \(F\) is the mid - point of \(DE\), \(DE = 2DF\). Substitute \(DE = 2DF\) into \(\frac{BD}{DE}=\frac{DC}{CE}\), we have \(\frac{BD}{2DF}=\frac{DC}{CE}\), which can be transformed into \(\frac{BD}{DF}=\frac{2DC}{CE}\). And because \(D\) is the mid - point of \(AC\), \(AC = 2DC\), so \(\frac{BD}{DF}=\frac{AC}{CE}\). Also, from \(Rt\triangle BDE\sim Rt\triangle DCE\), we know that \(\angle BDF=\angle ACE\) (because \(\angle BDE=\angle C\) and \(\angle BDF\) and \(\angle BDE\) are related, \(\angle ACE\) and \(\angle C\) are the same angle in the context of the triangle). According to the SAS (Side - Angle - Side) similarity criterion, \(\triangle BDF\sim\triangle ACE\).

Step3: Analyze angle relationships

From \(\triangle BDF\sim\triangle ACE\), we can get \(\angle DBF=\angle CAE\). Let the intersection of \(AE\) and \(BF\) be \(G\). In quadrilateral \(ADGB\), \(\angle DBF=\angle CAE\), and \(\angle ADB = 90^{\circ}\) (because \(BD\perp AC\)). The sum of the interior angles of a quadrilateral is \(360^{\circ}\), and we can also use the property of cyclic quadrilaterals. Since \(\angle DBF=\angle CAE\), the points \(A\), \(D\), \(G\), \(B\) are concyclic (a quadrilateral with an exterior angle equal to the interior opposite angle is cyclic). Then \(\angle BGA=\angle BDA = 90^{\circ}\) (angles in the same segment of a circle are equal).

Answer:

We have proved that \(AE\perp BF\) by using the properties of similar triangles and cyclic quadrilaterals. The key steps are proving the similarity of right - angled triangles, using the mid - point property to establish the similarity of another pair of triangles, and then using the angle relationships and the property of cyclic quadrilaterals to show that the angle between \(AE\) and \(BF\) is \(90^{\circ}\).