Two Concyclic Quadruples

$\angle ABC = 180^\circ - 20^\circ - 75^\circ = 85^\circ$ (angles in a triangle). Since we are given that $BDPF$ is concylic, it follows that $\angle FPD = 95^\circ$ (opposite angles of a cyclic quad)

Using the property that opposite angles of a cyclic quadrilateral sum up to 180 degrees, we have

<BAC + <FPE=180 (since AFPE is a cyclic quadrilateral)

This gives <FPE=160 deg

Similarly, <ACB + <DPE= 180 (since CDPE is a cyclic quadrilateral)

This gives <DPE=105 deg

Now the sum of angles around a point must be 360 degrees. Therefore,

<DPF= 360- (<DPE+ <FPE) = 360- (105+ 160) = 95 degrees

sum of any two opposite angles of circumscribed quadrilateral is 180 degree. ∠FAE+ ∠FPE =180; ∠FAE = ∠BAC =20; so ∠FPE =160; Similarly ∠DPE =105; but ∠DPE + ∠FPE + ∠DPF= 360; So, ∠DPF = 360- 160-105= 95 degree.

BAC is 20 degrees, therefore EPF will be (180-20)=160 degrees. similarly DPE will be 105 degrees. DPE + EPF + DPF=360. Equating these, DPF is 95 degrees.

We are given a triangle ABC in which $\angle BAC = 20 ^{\circ}$ and $\angle ACB = 75 ^{\circ}$ . Points D,E and F lie on BC,CA and AB, respectively, such that A,E,P,F are concyclic and C,D,P,E are concyclic. Now we draw PF, PE and PD. So, quadrilaterals AEPF and CDPE are cyclic. So the sum of their opposite angles is $180 ^{\circ}$ . Therefore,

$\angle FAE + \angle FPE = 180^{\circ}$ $\Rightarrow \angle FPE = 180^{\circ} - 20^{\circ} = 160^{\circ}$

Similarly, $\angle DPE = 105^{\circ}$

We know that the angle around the point $P$ is $360^{\circ}$ . So,

$\angle DPF + \angle FPE + \angle DPE = 360^{\circ}$ $\Rightarrow \angle DPF = 360^{\circ} - 160^{\circ} - 105^{\circ}$ $\Rightarrow \angle DPF = 95^{\circ}$

We use that four concyclic points are equivalent to a cyclic quadrilateral.

In a cyclic quadrilateral, the sum of opposite angles are 180°. Therefore, $\angle EPF=180°-20°=160°$ and $\angle DPE=180°-75°=105°$ .

Now, note that $\angle EPF$ , $\angle DPE$ and $\angle DPF$ form a full angle. Therefore, $\angle DPF=360°-160°-105°=95°$ , which answers our question.

Extend line $EP$ beyond $P$ to a point $Q$ . Note that $\angle QPD=\angle ECD=75^\circ$ , and $\angle QPF=\angle EAF=20^\circ$ . Then we have $\angle DPF=\angle QPD+\angle QPF=75^\circ +20^\circ=95^\circ$ , as desired.

More generally, we see that $\angle DPF=\angle BAC+\angle BCA=180^\circ -\angle ABC$ .

A concyclic quadrilateral has opposite angles such that there sum is 180.

Now, A,E,P,F are concyclic. So,

angle FAE + angle FPE= 180

20+ angle FPE = 180

angle FPE = 160

similarly, C,D,P,E are concyclic.

So,

angle DCE +angle DPE = 180

since angle DCE equals 75, angle DPE will be equal to 105 degree.

So, angle FPE +angle DPE=160+105=265

implies that reflex angle DPF =265

So angle DPF=360-265 =95

The tricky part is drawing the picture. You will need to insert line segment EP.

Angle FAE is an inscribed angle. Since it is 20 degrees, minor arc EF = 40 degrees, making major arc EF = 320 degrees; therefore, angle FPE = 160 degrees.

Similarly, angle DCE is an inscribed angle. Since it is 75 degrees, minor arc DE = 150 degrees, making major arc DE = 210 degrees and angle DPE = 105 degrees.

Angle FPE + Angle DPE + Angle FPD = 360 degrees 160 + 105 + x = 360 x = 95 Angle FPD = 95 degrees

angle epf = 180 -20= 160. angle epd = 180-75=105 angle dpf = 360-160-105=95

Since AEPF is cyclic quadrilateral we have EAF+ EPF then we get EPF=160, similarly we get EPD=105.

But, DPF=360-EPF-EPD=95

Given that A,E,P,F are concyclic.

So, * EAF+EPF=180 * or, EPF=180-20=160

By same method we get EPD=105

So ,DPF =360-105-160= 95

So,here is the answer 95

In Concyclic Quadrilaterals sum of opposite angles is 180.