Question

ABCD is a square.E,F are points on BC,CD such that <EAF=45°,<EAB=15°. BD intersects AE at P. What is the value of <PFC in degree.

Answer 1

Answer on Question #84157 – Math – Geometry

Question

ABCD is a square. E, F are points on BC, CD such that $\angle EAF = 45{}^\circ$, $\angle EAB = 15{}^\circ$. BD intersects AE at P. What is the value of $\angle PFC$ in degrees?

Solution

$\angle ABP = \angle DBC = 45{}^\circ$ because BD is the bisector of $\angle ABC$ ($\angle ABC = 90{}^\circ$ because ABCD is a square).

$\angle APB = 180{}^\circ - \angle BAP - \angle ABP = 180{}^\circ - 15{}^\circ - 45{}^\circ = 120{}^\circ$ (by theorem about the sum of angles of a triangle).

$\Delta ABP = \Delta BPC$ by the SAS theorem (PB is common, AB=BC as the sides of the square and the angle between them), then $\angle BPC = \angle APB = 120{}^\circ$.

$\angle BPC$ and $\angle DPC$ are adjacent angles, then

$$\angle CPD = 180{}^\circ - \angle BPC = 180{}^\circ - 120{}^\circ = 60{}^\circ.$$

$\Delta DPC = \Delta DPA$ by SAS theorem (AD=DC, $\angle PDC = \angle PDA$, PD is common), then

$$\angle PCD = \angle PAD = \angle DAF + \angle FAP = 30{}^\circ + 45{}^\circ = 75{}^\circ.$$

The triangles $\Delta AOP$ and $\Delta DOF$ are similar ($\angle AOP = \angle DOF$ as vertical, $\angle OAP = \angle ODF = 45{}^\circ$), then $\frac{OF}{OP} = \frac{OD}{OA}$.

The $\Delta ADO$ and the $\Delta PFO$ are similar triangles ($\angle AOD$ and $\angle POF$ are equal as vertical, $\frac{OF}{OP} = \frac{OD}{OA}$ because the triangles $\Delta AOP$ and $\Delta DOF$ are similar) then $\angle OFP = \angle ODA = 45{}^\circ$, in other words, $\angle AFP = \angle ADB = 45{}^\circ$.

$\angle AOD = 180{}^\circ - \angle DAO - \angle ADO = 180{}^\circ - 30{}^\circ - 45{}^\circ = 105{}^\circ$ (by theorem about the sum of angles of a triangle)

$$\begin{array}{l} \angle FOD = 180{}^\circ - \angle AOD = 180{}^\circ - 105{}^\circ = 75{}^\circ \\ \angle AFD = 180{}^\circ - \angle DOF - \angle ODF = 180{}^\circ - 75{}^\circ - 45{}^\circ = 60{}^\circ \\ \angle PFD = \angle PFO + \angle OFD = 45{}^\circ + 60{}^\circ = 105{}^\circ. \\ \angle PFC = 180{}^\circ - \angle PFD = 180{}^\circ - 105{}^\circ = 75{}^\circ. \end{array}$$

Answer: $\angle PFC = 75{}^\circ$.

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