Question
In Fig. $\angle\text{Q}>\angle\text{R}, PA$ is the bisector of $\angle\text{QPR}$ and $\text{PM}\perp\text{QR}$ Prove that $\angle\text{APM}=\frac{1}{2}(\angle\text{Q}-\angle\text{R}).$
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Answer
Given In $\Delta\text{PQR},\angle\text{Q}>\angle\text{R},$PA is the bisector of $\angle\text{QPR}$ and $\text{PM}\perp\text{QR}.$ To prove $\angle\text{APM}=\frac{1}{2}(\angle\text{Q}-\angle\text{R})$ Proof Since, $PA$ is the bisector of $\angle\text{QPR}.$
$\therefore\ \angle\text{QPA}=\angle\text{APR}$ In $\Delta\text{PQM},$
$\angle\text{PQM}+\angle\text{PMQ}+\angle\text{QPM}=180^\circ...(\text{i}) [$by angle sum property of a triangle$]$ $\Rightarrow\angle\text{PQM}+90^\circ+\angle\text{QPM}=180^\circ$
$\big[\because\ \text{PM}\perp\text{QR}\Rightarrow\angle\text{PMQ}=90^\circ\big]$
$\Rightarrow\angle\text{PQM}=90^\circ-\angle\text{QPM}...(\text{ii})$ In $\Delta\text{PMR},$
$\angle\text{PMR}+\angle\text{PRM}+\angle\text{RPM}=180^\circ [$by angle sum property of a triangle$]$
$\Rightarrow90^\circ+\angle\text{PRM}+\angle\text{RPM}=180^\circ$
$\big[\because\ \text{PM}\perp\text{QR}\Rightarrow\angle\text{PMR}=90^\circ\big]$
$\Rightarrow\angle\text{PRM}=180^\circ-90^\circ-\angle\text{RPM}$
$\Rightarrow\angle\text{PRM}=90^\circ-\angle\text{RPM}..(\text{iii})$ On subtracting Eq. $(iii)$ from Eq. $(ii),$
we get $\angle\text{Q}-\angle\text{R}=(90^\circ-\angle\text{QPM})-(90^\circ-\angle\text{RPM})$
$\big[$ where $\angle\text{PQM}=\angle\text{Q}$ and $\angle\text{PRM}=\angle\text{R}\big]$
$\Rightarrow\angle\text{Q}-\angle\text{R}=\angle\text{RPM}-\angle\text{QPM}$
$\Rightarrow\angle\text{Q}-\angle\text{R}=\big[\angle\text{RPA}+\angle\text{APM}\big]-\big[\angle\text{QPA}-\angle\text{APM}\big]..(\text{iv})$
$\Rightarrow\angle\text{Q}-\angle\text{R}=\angle\text{QPAS}+\angle\text{APM}-\angle\text{QPA}+\angle\text{APM} [$by using Eq. $(i)]$
$\Rightarrow\angle\text{Q}-\angle\text{R}=2\angle\text{APM}$
$\therefore\angle\text{APM}=\frac{1}{2}(\angle\text{Q}-\angle\text{R})$
$\therefore\ \angle\text{QPA}=\angle\text{APR}$ In $\Delta\text{PQM},$
$\angle\text{PQM}+\angle\text{PMQ}+\angle\text{QPM}=180^\circ...(\text{i}) [$by angle sum property of a triangle$]$ $\Rightarrow\angle\text{PQM}+90^\circ+\angle\text{QPM}=180^\circ$
$\big[\because\ \text{PM}\perp\text{QR}\Rightarrow\angle\text{PMQ}=90^\circ\big]$
$\Rightarrow\angle\text{PQM}=90^\circ-\angle\text{QPM}...(\text{ii})$ In $\Delta\text{PMR},$
$\angle\text{PMR}+\angle\text{PRM}+\angle\text{RPM}=180^\circ [$by angle sum property of a triangle$]$
$\Rightarrow90^\circ+\angle\text{PRM}+\angle\text{RPM}=180^\circ$
$\big[\because\ \text{PM}\perp\text{QR}\Rightarrow\angle\text{PMR}=90^\circ\big]$
$\Rightarrow\angle\text{PRM}=180^\circ-90^\circ-\angle\text{RPM}$
$\Rightarrow\angle\text{PRM}=90^\circ-\angle\text{RPM}..(\text{iii})$ On subtracting Eq. $(iii)$ from Eq. $(ii),$
we get $\angle\text{Q}-\angle\text{R}=(90^\circ-\angle\text{QPM})-(90^\circ-\angle\text{RPM})$
$\big[$ where $\angle\text{PQM}=\angle\text{Q}$ and $\angle\text{PRM}=\angle\text{R}\big]$
$\Rightarrow\angle\text{Q}-\angle\text{R}=\angle\text{RPM}-\angle\text{QPM}$
$\Rightarrow\angle\text{Q}-\angle\text{R}=\big[\angle\text{RPA}+\angle\text{APM}\big]-\big[\angle\text{QPA}-\angle\text{APM}\big]..(\text{iv})$
$\Rightarrow\angle\text{Q}-\angle\text{R}=\angle\text{QPAS}+\angle\text{APM}-\angle\text{QPA}+\angle\text{APM} [$by using Eq. $(i)]$
$\Rightarrow\angle\text{Q}-\angle\text{R}=2\angle\text{APM}$
$\therefore\angle\text{APM}=\frac{1}{2}(\angle\text{Q}-\angle\text{R})$
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