Question

$\angle ape$ and $\angle epd$ are congruent. find the measure of major arc $\overarc{edb}$.
draw
circle with center p, points a, b, c, d, e on circumference; angles at p: $\angle apb = 46^\circ$, $\angle bpc = 85^\circ$, $\angle cpd = 87^\circ$, and $\angle ape \cong \angle epd$

Explanation:

Step1: Find measure of ∠EPD

Since ∠APE ≅ ∠EPD, and from the diagram, the angle adjacent to ∠APE (assuming vertical or congruent) – wait, actually, the sum of central angles around a point is \(360^\circ\). First, note that ∠APE and ∠EPD are congruent. Let's find the measure of ∠EPD. Wait, maybe we can find the measure of ∠EPD by first finding the measure of ∠APE. Wait, the central angles: let's list the given angles: \(46^\circ\) (∠APB), \(85^\circ\) (∠BPC), \(87^\circ\) (∠CPD), and we need to find ∠DPE and ∠EPA (which are congruent). Let the measure of ∠APE = ∠EPD = \(x\). Then, sum of all central angles: \(46 + 85 + 87 + x + x = 360\). So, \(218 + 2x = 360\). Then, \(2x = 360 - 218 = 142\), so \(x = 71^\circ\). So ∠EPD = \(71^\circ\), ∠APE = \(71^\circ\). Now, we need to find the measure of major arc \( \overarc{EDB} \). The major arc EDB: let's find the sum of the central angles corresponding to arc EDB. Arc EDB goes from E to D to B. So the central angles for arc EDB would be ∠EPD + ∠DPC + ∠CPB + ∠BPA? Wait, no. Wait, major arc: the major arc EDB would be the longer path from E to B through D. Wait, let's check the central angles. Wait, the central angles: from E to D is ∠EPD = \(71^\circ\), D to C is ∠CPD = \(87^\circ\), C to B is ∠BPC = \(85^\circ\), B to A is ∠APB = \(46^\circ\)? No, that's not right. Wait, maybe I made a mistake. Wait, the major arc EDB: the minor arc EDB would be the shorter path, so major arc is \(360^\circ\) minus minor arc EDB. Wait, no, let's find the central angle for major arc EDB. Wait, the points are E, D, B. So the central angle for arc EDB is the sum of the central angles from E to D to B. Wait, E to D: ∠EPD = \(71^\circ\), D to B: let's see, from D to B, what's the central angle? From D to C is \(87^\circ\), C to B is \(85^\circ\), so D to B is \(87 + 85 = 172^\circ\)? No, that's not. Wait, maybe I messed up the direction. Wait, let's list all central angles:

- ∠APB = \(46^\circ\) (A to B)
- ∠BPC = \(85^\circ\) (B to C)
- ∠CPD = \(87^\circ\) (C to D)
- ∠DPE = \(71^\circ\) (D to E)
- ∠EPA = \(71^\circ\) (E to A)

Now, major arc EDB: starts at E, goes through D, to B. So the central angle for major arc EDB is ∠EPD + ∠DPC + ∠CPB? Wait, E to D is ∠EPD = \(71^\circ\), D to C is ∠CPD = \(87^\circ\), C to B is ∠BPC = \(85^\circ\), B to A is ∠APB = \(46^\circ\)? No, that's E to D to C to B to A, which is too long. Wait, no, major arc EDB: the major arc from E to B through D. So the central angle for major arc EDB would be the sum of the central angles from E to D to B. Wait, E to D: ∠EPD = \(71^\circ\), D to B: let's see, from D to B, the central angle is ∠DPB. Wait, ∠DPB is ∠DPC + ∠CPB = \(87 + 85 = 172^\circ\). Then E to D is \(71^\circ\), D to B is \(172^\circ\)? No, that's not. Wait, maybe the major arc EDB's central angle is ∠EPB, but major arc. Wait, let's calculate the measure of major arc EDB by summing the central angles that make up the major arc. The major arc EDB will pass through D, C, B? Wait, no, let's find the measure of the major arc EDB. The central angle for major arc EDB: let's see, the minor arc EDB would be the central angle ∠EPB. Wait, ∠EPB: from E to P to B. So ∠EPB is ∠EPA + ∠APB = \(71 + 46 = 117^\circ\). So minor arc EDB has central angle \(117^\circ\), so major arc EDB is \(360 - 117 = 243^\circ\)? Wait, that can't be. Wait, no, maybe I messed up the central angles. Wait, let's re-express:

Given that ∠APE = ∠EPD = \(71^\circ\) (from earlier calculation: \(46 + 85 + 87 + 2x = 360\) → \(218 + 2x = 360\) → \(2x = 142\) → \(x = 71\)). So ∠APE = \(71^\circ\), ∠EPD = \(71^\circ\).

Now, the central angle for arc EDB (minor arc) would be ∠EPB. ∠EPB is ∠EPA + ∠APB = \(71 + 46 = 117^\circ\). Therefore, the major arc EDB would be \(360 - 117 = 243^\circ\)? Wait, but that seems low. Wait, no, maybe the major arc EDB is the sum of the central angles from E to D to B, which is ∠EPD + ∠DPC + ∠CPB. Wait, ∠EPD = \(71^\circ\), ∠DPC = \(87^\circ\), ∠CPB = \(85^\circ\). So \(71 + 87 + 85 = 243^\circ\). Wait, that's the same as before. Wait, but let's check:

Wait, the central angles:

- ∠APE = \(71^\circ\) (E to A)
- ∠APB = \(46^\circ\) (A to B)
- ∠BPC = \(85^\circ\) (B to C)
- ∠CPD = \(87^\circ\) (C to D)
- ∠DPE = \(71^\circ\) (D to E)

Now, major arc EDB: from E to D to B. So the central angles are D to E? No, E to D is \(71^\circ\) (∠EPD), D to C is \(87^\circ\) (∠CPD), C to B is \(85^\circ\) (∠BPC). So sum: \(71 + 87 + 85 = 243^\circ\). Wait, but that's the same as \(360 - 117\) (since minor arc EDB is \(117^\circ\)). So major arc EDB is \(243^\circ\). Wait, but let's verify the initial calculation of ∠APE and ∠EPD. The sum of all central angles: \(71\) (∠APE) + \(46\) (∠APB) + \(85\) (∠BPC) + \(87\) (∠CPD) + \(71\) (∠DPE) = \(71 + 46 = 117\), \(117 + 85 = 202\), \(202 + 87 = 289\), \(289 + 71 = 360\). Perfect, that sums to \(360^\circ\), so that's correct. Then, the major arc EDB: the central angle for major arc EDB is the sum of ∠EPD + ∠CPD + ∠BPC = \(71 + 87 + 85 = 243^\circ\). So the measure of major arc \( \overarc{EDB} \) is \(243^\circ\).

Step1: Calculate ∠APE and ∠EPD

Let \( \angle APE = \angle EPD = x \). Sum of central angles:
\( 46^\circ + 85^\circ + 87^\circ + x + x = 360^\circ \)
Simplify: \( 218^\circ + 2x = 360^\circ \)
Solve: \( 2x = 360^\circ - 218^\circ = 142^\circ \) ⟹ \( x = 71^\circ \).

Step2: Find central angle for minor arc \( \overarc{EDB} \)

Minor arc \( \overarc{EDB} \) corresponds to \( \angle EPB \):
\( \angle EPB = \angle APE + \angle APB = 71^\circ + 46^\circ = 117^\circ \).

Step3: Calculate major arc \( \overarc{EDB} \)

Major arc measure = \( 360^\circ - \) minor arc measure:
\( 360^\circ - 117^\circ = 243^\circ \).

Answer:

\(243^\circ\)