(I) In\[\vartriangle \text{ }ABC\text{ }and\text{ }\vartriangle \text{ }AMP\], we have \[\angle BAC\text{ }=\angle PAM\text{ }\left[ Common \right]\] \[\angle ABC\text{ }=\text{ }\angle PMA\text{...
Answer: In \[\Delta \text{ }ADC\text{ }and\text{ }\Delta \text{ }BAC,\] \[\angle ADC\text{ }=\angle BAC\text{ }\left[ Given \right]\] \[\angle ACD\text{ }=\angle ACB\text{ }\left[ Common \right]\]...
Answer: In\[\Delta \text{ }DHG\text{ }and\text{ }\Delta \text{ }DFE\], \[\angle GHD\text{ }=\text{ }\angle DFE\text{ }=\text{ }{{90}^{o}}\] \[\angle D\text{ }=\angle D\text{ }\left[ Common \right]\]...
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(i) True (ii) True
(i) True (ii) False
(i) False (ii) True
In \[\Delta \text{ }ABC,\] \[AC\text{ }=\text{ }AB\text{ }\left[ Given \right]\] In this way, \[\angle ABC\text{ }=\angle ACB\][Angles inverse to rise to sides are equal.] In \[\Delta \text{...
Answer: In \[\Delta \text{ }DOQ\text{ }and\text{ }\Delta \text{ }BOP,\] \[\angle QDO\text{ }=\angle PBO\] \[\left[ As\text{ }AB\text{ }\left| \left| \text{ }DC\text{ }in\text{ }this\text{...
Answer: In \[\Delta \text{ }ABD\text{ }and\text{ }\Delta \text{ }CAE,\] \[\angle ADE\text{ }=\angle AED\text{ }\left[ Angles\text{ }inverse\text{ }to\text{ }rise\text{ }to\text{ }sides\text{...
Answer: (I) In\[\Delta \text{ }AME\text{ }and\text{ }\Delta \text{ }ANC\], \[\angle AME\text{ }=\angle ANC\] \[\left[ Since\text{ }DE\text{ }\left| \left| \text{ }BC\text{ }in\text{ }this\text{...
Answer: In \[\Delta \text{ }ABM\text{ }and\text{ }\Delta \text{ }ACN,\] \[\angle AMB\text{ }=\text{ }\angle ANC\] [Since, BM ⊥ AC and CN ⊥ AB] \[\angle BAM\text{ }=\angle CAN\] [Common angle] Thus,...
(I) In\[\Delta \text{ }ABC\], we have \[\angle ABC\text{ }=\text{ }2\angle ACB\][Given] Presently, let \[\angle ACB\text{ }=\text{ }x\] In this way, \[\angle ABC\text{ }=\text{ }2x\] Likewise given,...
(I) Given, \[AO\text{ }=\text{ }2CO\text{ }and\text{ }BO\text{ }=\text{ }2DO,\] \[AO/CO\text{ }=\text{ }2/1\text{ }=\text{ }BO/DO\] What's more, \[\angle AOB\text{ }=\angle DOC\] [Vertically inverse...
(I) As\[AD||BC\], we have \[AD||\text{ }BP\] too. Along these lines, by \[BPT\] \[DP/PL\text{ }=\text{ }AB/BL\] What's more, since \[ABCD\]is a parallelogram, \[AB\text{ }=\text{ }DC\] Consequently,...
(I) In\[\vartriangle APB\text{ }and\text{ }\vartriangle CPD\], we have \[\angle APB\text{ }=\angle CPD\] [Vertically inverse angles] \[\angle ABP\text{ }=\angle CDP\] [Alternate points as, AB||DC]...
Answer: (I) In \[\vartriangle \mathbf{APC}\text{ }\mathbf{and}\text{ }\vartriangle \mathbf{BPD}\]we have \[\angle APC\text{ }=\angle BPD\] [Vertically inverse angles] \[\angle ACP\text{ }=\angle...