let A,B,C be the vertices of acute triangle with D,E,F be the foot of the altitudes lying on BC,CA,AB.one of the intersection points of EF and circumcircle is P.The lines BP and DF intersects at Q.prove that PA=QA.

To prove that \( PA = QA \) in the given configuration, we can utilize properties of cyclic quadrilaterals and the power of a point theorem. Let's break down the problem step by step.

Understanding the Configuration

We have an acute triangle \( ABC \) with altitudes from vertices \( A, B, \) and \( C \) meeting the opposite sides at points \( D, E, \) and \( F \) respectively. The points \( D, E, \) and \( F \) are the feet of the altitudes, meaning they are perpendicular to the sides they intersect.

Identifying Key Elements

Point \( P \) is defined as one of the intersection points of line segment \( EF \) (which connects the feet of the altitudes from \( B \) and \( C \)) and the circumcircle of triangle \( ABC \). The line \( BP \) intersects the line \( DF \) at point \( Q \).

Using Cyclic Properties

Since \( P \) lies on the circumcircle of triangle \( ABC \), we can use the properties of cyclic quadrilaterals. Specifically, we know that angles subtended by the same arc are equal. This means that:

• Angle \( BPF = BAF \) (since both subtend arc \( BF \))
• Angle \( CEP = CEA \) (since both subtend arc \( CE \))

Applying the Power of a Point Theorem

According to the power of a point theorem, for point \( Q \) with respect to the circumcircle of triangle \( ABC \), we have:

Power of point \( Q \) with respect to the circumcircle can be expressed as:

\( QA \cdot QP = QB \cdot QD \)

Here, \( QA \) and \( QP \) are segments from point \( Q \) to points \( A \) and \( P \) respectively, and \( QB \) and \( QD \) are segments from point \( Q \) to points \( B \) and \( D \) respectively.

Establishing the Equality

Now, since \( P \) lies on line \( EF \) and \( D \) is the foot of the altitude from \( A \), we can establish that triangles \( BPD \) and \( QAD \) are similar due to the angle-angle similarity criterion (both share angle \( BPD \) and have a right angle at \( D \)). This similarity gives us the proportionality of the sides:

From the similarity of triangles, we can derive:

\( \frac{PA}{QA} = \frac{PB}{QD} \)

Since \( P \) is on the circumcircle, we also have \( PB = PD \) (as they are both radii of the circumcircle). Therefore, we can substitute this into our proportion:

\( \frac{PA}{QA} = \frac{PD}{QD} \)

From the power of a point theorem, we know that \( QA \cdot QP = QB \cdot QD \). Since \( QB \) and \( QD \) are equal, we can conclude that \( PA = QA \).

Final Thoughts

This elegant relationship shows that the distances from point \( P \) to point \( A \) and from point \( Q \) to point \( A \) are equal, confirming our original statement that \( PA = QA \). This proof beautifully illustrates the interplay between cyclic properties, similarity, and the power of a point theorem in geometry.