Question

P is the centre of a circle. Chord MN = chord ML. Also, PQ is perpendicular to MN and PR is perpendicular ML. Prove that chord QN = chord RL.
​

Answer 1

Proved that QN = RL if  P is the centre of a circle. Chord MN = chord ML and PQ ⊥ MN , PR ⊥ML

Step-by-step explanation:

in Δ MPX & NPX

MP = NP  ( Radius)

PX = PX  ( common)

∠MXP = ∠NXP = 90°

=> Δ MPX ≅ NPX

=> MX = NX

=> MX + NX = MN

=> NX = MN/2

Simialrly

Δ MPY ≅ LPY    

=> LY = ML/2

ML = MN

=> ML/2 = MN/2

=> MY = YL = NX = MX

PX² = MP² - MX²  & PY² = PM² - MY²

=> PX = PY  

QX = QP - PX

YR = RP - PY

QP = RP  ( Radius)  & PX = PY

=> QX  = YR

now in Δ QXN  & ΔRYL

QN² = QX² + NX²

RL² =  YR² + YL²

QX  = YR & YL = NX

=> QN² = RL²

=> QN = RL

QED

Proved

Learn More

A chord AB of a circle (0,r) is produced to P

https://brainly.in/question/12832925

Prove that two opposite arcs formed

https://brainly.in/question/13449747

Answer 2

PROVED

Step-by-step explanation:

here , in triangle MPX and NPX

MP = PM (radii)

PX = XP (common)

$\angle MXP = \angle NXP = 90^\circ$

therefore ,

$\bigtriangleup MPX \cong \bigtriangleup NPX \\\\MX = NX \\\\MX + NX = MN \\\\\NX + NX = MN \\\\2NX = MN \\\\NX = \frac{MN}{2}\\\\\text{similarly }\\\\\bigtriangleup MPY \cong \bigtriangleup PLY \\\\LY = \frac{ML}{2}\\\\ML = MN \\\\\frac{ML}{2} = \frac{MN}{2}\\\\therefore , \\\\MY = LY = NX = XM$

now , in triangle XMP and triangle PMY

using pythagoras theorem

$PX^2 = MP^2-MX^2 , PY^2 = PM^2-MY^2 \\\\PX = PY \\\\QX= QP-PX\\\\YE=PR-PY\\\\QX = YR$

NOW , similarly in triangle QXN and LYR

using  PGT

$QN^2 = QX^2+NX^2 , RL^2 = YR^2-YL^2 \\\\QX = YR, YL = NX \\\\QN^2= RL^2 \\\\QN = RL$

HENCE PROVED

#Learn more :

In figure seg MN is a chord of a circle with center o MN =25 L is a point on chord MN such that ML=9and OL=5find the radius of circle​

https://brainly.in/question/14413384