Question

Plzzzzzzzzzz giveeeeeee ansssssssss

Answer 1

angle RPQ = a deg

Angle RQP = 90-a deg

as angle PRQ = 90 deg given

angle PDR &Angle QDR = 90 deg given

in Triangle PRD

tana = RD/PD => RD = PD Tana

in triangle QRD

tan (90-a) = RD/ DQ => RD = DQ Tan(90-a)

=> RD = DQ/Tana

equating both

PD Tan = DQ/ Tan a

=> PD/DQ = 1/Tan^2 a = LHS

RE is bisector of angle PRQ (90 deg)

so angle PRE = 45 deg angle QRE = 45 deg

In triangle PRE

RE/Sina = PE/Sin45

=> RE = PESina/ Sin45

in triangle QRE

RE/Sin(90-a) = EQ/Sin45

=> RE = EQ Sin(90-a)/ Sin45

= RE = EQ Cosa/ Sin45

equating RE

PE Sina/Sin45 = EQ Cosa/Sin 45

cancelling Sin 45

=>PE Sina = EQ Cosa

=> PE/EQ = Cosa / Sina

=> PE/ EQ = 1/Tana

squaring both sides

PE^2/EQ^2 = 1/Tan^2a = RHS

LHS = RHS

PD/DQ = PE^2/EQ^2