Question

IF PA=6,PB=3,CE=5
Find BD & PE​

Answer 1

Answer:

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Step-by-step explanation:

$to \: find = \\ BD \: \: and \: \: PE \\ \\ so \: here \\ \: PA=6,PB=3,CE=5 \\ \\ PA \: \: is \: a \: tangent \: and \: PB \: is \: a \: secant \\ so \: by \: applying \: \\ tangent \: secant \: theorem \\ PA {}^{2} = PB \times PD \\ (6) {}^{2} = 3 \times PD \\ \\ \: PD = \frac{36}{3} \\ \\ PD = 12 \\ so \: hence \: then \\ \: BD = PD - PB \\ = 12 - 6 \\ =6 .units \\ \\ BD = 6$

$moreover \: here \\ PD \: \: and \: \: PE \: are \: secant \: intersecting \: \\ externally \\ so \: by \: applying \\ \\external \: intersection \: of \: secants \: property \\ we \: get \\ PB×PD=PC×PE \\ 12 \times 3 =PC×(PC+CE) \: \\ 36 = PC(PC + 5) \\ = PC {}^{2} \: + \: 5PC \\ \\ PC {}^{2} + 5PC - 36 = 0 \\ \\ PC {}^{2} + 9PC - 4PC - 36 = 0 \\ \\ PC(PC + 9) - 4(9 + PC) = 0 \\ \\ (PC + 9)(PC - 4) = 0 \\ \\ PC = - 9 \: \: or \: \: PC = 4 \\ but \: as \: length \: is \: never \: negative \: \\ PC = - 9 \: \: is \: \: absurd \\ \\ hence \: then \\ PC = 4$

$so \: \: as \: \: (P-C-E) \\ PE=PC+CE \\ = 4 + 5 \\ = 9.units \\ \\ PE = 9$