Question

In Fig.44, it is given that LM=MN, QM=MR, ML⊥PQ and MN⊥PR. Prove that PQ=PR.​

Answer 1

Answer:

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Answer 2

${ \purple \implies} \huge \rm { \red{given \: that}}$

$\implies \large \rm{(lm = mn)} \\ \: \: \: \: \: \: \: \: \: \: \large \rm (qm = mr) \\ \: \: \: \: \: \large \rm{(ml \: is \: perpendicular \: to \: pq)} \\ \: \: \: \large \rm{(mn \: is \: perpendicular \: to \: pr)}$

$\purple \implies \huge \rm{ \red{to \: prove}}$

$\large \bold{pq = pr}$

$\: \: \: \: \: \: \: \: \: \: \: {\red \implies} { \rm {now \: proof}}$

$\rm{ \purple{in \: triangle \: qml \: and \: triangle \: rmn}}$

$\large \implies{ \huge \rm { \pink{ml = mn}}}$

$\large \implies{ \huge \rm { \pink{qm = mr}}}$

$\implies{ \rm { \pink{angle \: mlq = angle \: mnr}}}$

Hence,

∆QML=~∆RMN

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \large \rm{ \orange{by \: rhs \: criteria}}$

Now,

angle Q = angle R

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \large \rm { \purple{by \: cpct}}$

similarly, PQ=PR

$\: \: \: \: \: \: \: \: \: \large \frak { \red{hence \: proved}}$